Apollonius was born at Perga, in modern day Turkey. His greatest work was called “conics” which introduced curves like circle, parabola geometrically. He wrote six other books all related to the basics of modern day coordinate geometry.
His ideas were applied to study planetary theory and solve practical problems. He developed the sundial and contributed to other branches of science using his exceptional geometric skills. For this reason, Apollonius is hailed as “The Great Geometer”.
Coordinate geometry, also called Analytical geometry is a branch of mathematics, in which curves in a plane are represented by algebraic equations. For example, the equation $x^2+y^2=1$, describes a circle of unit radius in the plane. Thus coordinate geometry can be seen as a branch of mathematics which interlinks algebra and geometry, where algebraic equations are represented by geometric curves. This connection makes it possible to reformulate problems in geometry to problems in algebra and vice versa. Thus, in coordinate geometry, the algebraic equations have visual representations thereby making our understanding much deeper. For instance, the first degree equation in two variables $ax+by+c=0$ represents a straight line in a plane. Overall, coordinate geometry is a tool to understand concepts visually and created new branches of mathematics in modern times.
In the earlier classes, we initiated the study of coordinate geometry where we studied about coordinate axes, coordinate plane, plotting of points in a plane, distance between two points, section formulae, etc. All these concepts form the basics of coordinate geometry. Let us now recall some of the basic formulae.
In your earlier classes, you have studied how to calculate the area of a triangle when its base and corresponding height (altitude) are given. You have used the formula.
$$
\text{Area of triangle}=\frac12\times\text{base}\times\text{altitude}\text{ sq.units}.
$$
Fig. 5.6
With any three non-collinear points $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(y_3,y_3)$ on a plane, we can form a triangle $ABC$.
Using distance between two points formula, we can calculate $AB=c$, $BC=a$, $CA=b$. $a,b,c$ represent the lengths of the sides of the triangle $ABC$.
Using $2s=a+b+c$, we can calculate the area of triangle $ABC$ by using the Heron’s formula $\sqrt{s(s-a)(s-b)(s-c)}$. But this procedure of finding sides of $\triangle ABC$ and then calculating its area will be a tedious procedure.
There is an elegant way of finding area of a triangle using the coordinates of its vertices. We shall discuss such a method below.
Let $ABC$ be any triangle whose vertices are at $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$.
Draw $AP$, $BQ$ and $CR$ perpendiculars from $A$, $B$ and $C$ to the $x$-axis, respectively.
Clearly $ABQP$, $APRC$ and $BQRC$ are all trapeziums.
Fig. 5.7
Now from Fig. 5.7, it is clear that
$$
\text{Area of }\triangle ABC=\text{Area of trapezium }ABQP+\text{Area of trapezium }APRC-\text{Area of trapezium }BQRC.
$$
You also know that, the area of trapezium
$$
=\frac12\times\text{(sum of parallel sides)}\times\text{(perpendicular distance between the parallel sides)}.
$$
The vertices $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ of $\triangle ABC$ are said to be “taken in order” if $A,B,C$ are taken in anticlockwise direction. If we do this, then area of $\triangle ABC$ will never be negative.
If three distinct points $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ are collinear, then we cannot form a triangle, because for such a triangle there will be no altitude (height). Therefore, three points $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ will be collinear if the area of $\triangle ABC=0$.
Note — Another condition for collinearity
If $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ are collinear points, then
$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$
or $x_1y_2+x_2y_3+x_3y_1=x_1y_3+x_2y_1+x_3y_2$.
Similarly, if the area of $\triangle ABC$ is zero, then the three points lie on the same straight line. Thus, three distinct points $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ will be collinear if and only if area of $\triangle ABC=0$.
The following pictorial representation helps us to write the above formula very easily. Take the vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ and $D(x_4,y_4)$ in counter-clockwise direction and write them column-wise as that of the area of a triangle.
Pictorial multiplication pattern for the area of a quadrilateral
$$
\text{Area of the quadrilateral }ABCD=\frac12\{(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4)\}\text{ sq.units}.
$$
Note
To find the area of a quadrilateral, we divide it into triangular regions, which have no common area and then add the area of these regions.
The area of the quadrilateral is never negative. That is, always take the area of quadrilateral as positive.
Thinking Corner
If the area of a quadrilateral formed by the points $(a,a)$, $(-a,a)$, $(a,-a)$ and $(-a,-a)$, where $a\ne0$ is 64 square units, then identify the type of the quadrilateral.
The floor of a hall is covered with identical tiles which are in the shapes of triangles. One such triangle has the vertices at $(-3,2)$, $(-1,-1)$ and $(1,2)$. If the floor of the hall is completely covered by 110 tiles, find the area of the floor.
Solution Vertices of one triangular tile are at $(-3,2)$, $(-1,-1)$ and $(1,2)$.
$$
\begin{aligned}
\text{Area of this tile}&=\frac12\{(3-2+2)-(-2-1-6)\}\text{ sq.units}\\
&=\frac12(12)=6\text{ sq.units}.
\end{aligned}
$$
Since the floor is covered by 110 triangle shaped identical tiles,
$$
\text{Area of floor}=110\times6=660\text{ sq.units}.
$$
The given diagram shows a plan for constructing a new parking lot at a campus. It is estimated that such construction would cost ₹1300 per square feet. What will be the total cost for making the parking lot?
Solution The parking lot is a quadrilateral whose vertices are at $A(2,2)$, $B(5,5)$, $C(4,9)$ and $D(1,7)$.
Fig. 5.14
Pictorial multiplication pattern for the parking lot area
Find the area of the triangle formed by the points:
(i) $(1,-1)$, $(-4,6)$ and $(-3,-5)$
(ii) $(-10,-4)$, $(-8,-1)$ and $(-3,-5)$
Determine whether the sets of points are collinear?
(i) $\left(-\dfrac12,3\right)$, $(-5,6)$ and $(-8,8)$
(ii) $(a,b+c)$, $(b,c+a)$ and $(c,a+b)$
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘$p$’.
S.No.
Vertices
Area (sq.units)
(i)
$(0,0),(p,8),(6,2)$
20
(ii)
$(p,p),(5,6),(5,-2)$
32
In each of the following, find the value of ‘$a$’ for which the given points are collinear.
(i) $(2,3)$, $(4,a)$ and $(6,-3)$
(ii) $(a,2-2a)$, $(-a+1,2a)$ and $(-4-a,6-2a)$
Find the area of the quadrilateral whose vertices are at
(i) $(-9,-2)$, $(-8,-4)$, $(2,2)$ and $(1,-3)$
(ii) $(-9,0)$, $(-8,6)$, $(-1,-2)$ and $(-6,-3)$
Find the value of $k$, if the area of a quadrilateral is 28 sq.units, whose vertices are taken in the order $(-4,-2)$, $(-3,k)$, $(3,-2)$ and $(2,3)$.
If the points $A(-3,9)$, $B(a,b)$ and $C(4,-5)$ are collinear and if $a+b=1$, then find $a$ and $b$.
Let $P(11,7)$, $Q(13.5,4)$ and $R(9.5,4)$ be the mid-points of the sides $AB$, $BC$ and $AC$ respectively of $\triangle ABC$. Find the coordinates of the vertices $A$, $B$ and $C$. Hence find the area of $\triangle ABC$ and compare this with area of $\triangle PQR$.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.
Quadrilateral swimming pool surrounded by concrete patio
A triangular shaped glass with vertices at $A(-5,-4)$, $B(1,6)$ and $C(7,-4)$ has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
In the figure, find the area of (i) triangle $AGF$ (ii) triangle $FED$ (iii) quadrilateral $BCEG$.
Triangles AGF and FED and quadrilateral BCEG on a coordinate grid
The inclination of a line or the angle of inclination of a line is the angle which a straight line makes with the positive direction of $X$ axis measured in the counter-clockwise direction to the part of the line above the $X$ axis. The inclination of the line is usually denoted by $\theta$.
Note
The inclination of $X$ axis and every line parallel to $X$ axis is $0^\circ$.
The inclination of $Y$ axis and every line parallel to $Y$ axis is $90^\circ$.
While laying roads one must know how steep the road will be. Similarly, when constructing a staircase, we should consider its steepness. For the same reason, anyone travelling along a hill or a bridge, feels hard compared to traveling along a plain road.
All these examples illustrate one important aspect called “Steepness”. The measure of steepness is called slope or gradient.
The concept of slope is important in economics because it is used to measure the rate at which the demand for a product changes in a given period of time on the basis of its price. Slope comprises of two factors namely steepness and direction.
Fig. 5.16
Definition
If $\theta$ is the angle of inclination of a non-vertical straight line, then $\tan\theta$ is called the slope or gradient of the line and is denoted by $m$.
Therefore the slope of the straight line is $m=\tan\theta$, $0\leq\theta\leq180^\circ$, $\theta\neq90^\circ$.
To find the slope of a straight line when two points are given#
Two non-vertical lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1m_2=-1$.
Let $l_1$ and $l_2$ be two non-vertical lines with slopes $m_1$ and $m_2$, respectively. Let their inclinations be $\theta_1$ and $\theta_2$ respectively.
Then $m_1=\tan\theta_1$ and $m_2=\tan\theta_2$.
First we assume that, $l_1$ and $l_2$ are perpendicular to each other.
Fig. 5.22
Then $\angle ABC=90^\circ-\theta_1$ (sum of angles of $\triangle ABC$ is $180^\circ$).
Now measuring slope of $l_2$ through angles $\theta_2$ and $90^\circ-\theta_1$, which are opposite to each other, we get
The line $r$ passes through the points $(-2,2)$ and $(5,8)$ and the line $s$ passes through the points $(-8,7)$ and $(-2,0)$. Is the line $r$ perpendicular to $s$?
Solution
The slope of line $r$ is
$$
m_1=\frac{8-2}{5+2}=\frac{6}{7}
$$
The slope of line $s$ is
$$
m_2=\frac{0-7}{-2+8}=\frac{-7}{6}
$$$$
\text{The product of slopes}=\frac{6}{7}\times\frac{-7}{6}=-1
$$
That is, $m_1m_2=-1$.
Therefore, the line $r$ is perpendicular to line $s$.
Show that the points $(-2,5)$, $(6,-1)$ and $(2,2)$ are collinear.
Solution
The vertices are $A(-2,5)$, $B(6,-1)$ and $C(2,2)$.
$$
\begin{aligned}
\text{Slope of }AB&=\frac{-1-5}{6+2}=\frac{-6}{8}=\frac{-3}{4}\\
\text{Slope of }BC&=\frac{2+1}{2-6}=\frac{3}{-4}=\frac{-3}{4}
\end{aligned}
$$
We get, Slope of $AB=$ Slope of $BC$.
Therefore, the points $A$, $B$, $C$ all lie in a same straight line.
Without using Pythagoras theorem, show that the points $(1,-4)$, $(2,-3)$ and $(4,-7)$ form a right angled triangle.
Solution
Let the given points be $A(1,-4)$, $B(2,-3)$ and $C(4,-7)$.
$$
\begin{aligned}
\text{The slope of }AB&=\frac{-3+4}{2-1}=\frac{1}{1}=1\\
\text{The slope of }BC&=\frac{-7+3}{4-2}=\frac{-4}{2}=-2\\
\text{The slope of }AC&=\frac{-7+4}{4-1}=\frac{-3}{3}=-1
\end{aligned}
$$$$
\text{Slope of }AB\times\text{slope of }AC=(1)(-1)=-1
$$
$AB$ is perpendicular to $AC$. $\angle A=90^\circ$.
Therefore, $\triangle ABC$ is a right angled triangle.
Thinking Corner
Provide three examples of using the concept of slope in real-life situations.
Prove analytically that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to half of its length.
Solution
Let $P(a,b)$, $Q(c,d)$ and $R(e,f)$ be the vertices of a triangle.
Let $S$ be the mid-point of $PQ$ and $T$ be the mid-point of $PR$.
Any first degree equation in two variables $x$ and $y$ of the form $ax+by+c=0$ …(1) where $a$, $b$, $c$ are real numbers and at least one of $a$, $b$ is non-zero is called “Straight line” in $XY$ plane.
The $X$ axis and $Y$ axis together are called coordinate axes. The $x$ coordinate of every point on $OY$ ($Y$ axis) is 0. Therefore equation of $OY$ ($Y$ axis) is $x=0$ (fig 5.27).
Fig. 5.27
The $y$ coordinate of every point on $OX$ ($X$ axis) is 0. Therefore the equation of $OX$ ($X$ axis) is $y=0$ (fig 5.28).
Fig. 5.28
5.5.2 Equation of a straight line parallel to $X$ axis#
Let $AB$ be a straight line parallel to $X$ axis, which is at a distance ‘$b$’. Then $y$ coordinate of every point on ‘$AB$’ is ‘$b$’. (fig 5.29)
Fig. 5.29
Therefore, the equation of $AB$ is $y=b$.
Note
If $b>0$, then the line $y=b$ lies above the $X$ axis.
If $b<0$, then the line $y=b$ lies below the $X$ axis.
If $b=0$, then the line $y=b$ is the $X$ axis itself.
5.5.3 Equation of a Straight line parallel to the $Y$ axis#
Let $CD$ be a straight line parallel to $Y$ axis, which is at a distance ‘$c$’. Then $x$ coordinate of every point on $CD$ is ‘$c$’. The equation of $CD$ is $x=c$. (fig 5.30)
Fig. 5.30
Note
If $c>0$, then the line $x=c$ lies right to the side of the $Y$ axis.
If $c<0$, then the line $x=c$ lies left to the side of the $Y$ axis.
If $c=0$, then the line $x=c$ is the $Y$ axis itself.
Every straight line that is not vertical will cut the $Y$ axis at a single point. The $y$ coordinate of this point is called $y$ intercept of the line.
A line with slope $m$ and $y$ intercept $c$ can be expressed through the equation
$$
y=mx+c
$$
We call this equation as the slope-intercept form of the equation of a line.
Do You Know?
If a line with slope $m$, $m\ne0$ makes $x$ intercept $d$, then the equation of the straight line is $y=m(x-d)$.
$y=mx$ represent equation of a straight line with slope $m$ and passing through the origin.
The graph relates temperatures $y$ (in Fahrenheit degree) to temperatures $x$ (in Celsius degree) (a) Find the slope and $y$ intercept (b) Write an equation of the line (c) What is the mean temperature of the earth in Fahrenheit degree if its mean temperature is $25^\circ$ Celsius?
Solution
(a) From the figure,
$$
\text{slope}=\frac{\text{change in }y\text{ coordinate}}{\text{change in }x\text{ coordinate}}=\frac{68-32}{20-0}=\frac{36}{20}=\frac95=1.8
$$
The line crosses the $Y$ axis at $(0,32)$.
So the slope is $\dfrac95$ and $y$ intercept is 32.
Fig. 5.31
(b) Use the slope and $y$ intercept to write an equation.
The equation is $y=\dfrac95x+32$.
(c) In Celsius, the mean temperature of the earth is $25^\circ$. To find the mean temperature in Fahrenheit, we find the value of $y$ when $x=25$.
Therefore, the mean temperature of the earth is $77^\circ\mathrm{F}$.
Note
The formula for converting Celsius to Fahrenheit is given by $F=\dfrac95C+32$ which is the linear equation representing a straight line derived in the example.
Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be two given distinct points. Slope of the straight line passing through these points is given by $m=\dfrac{y_2-y_1}{x_2-x_1}$, $(x_2\ne x_1)$.
From the equation of the straight line in point slope form, we get
Two buildings of different heights are located at opposite sides of each other. If a heavy rod is attached joining the terrace of the buildings from $(6,10)$ to $(14,12)$, find the equation of the rod joining the buildings?
Solution Let $A(6,10)$, $B(14,12)$ be the points denoting the terrace of the buildings.
The equation of the rod is the equation of the straight line passing through $A(6,10)$ and $B(14,12)$.
We will find the equation of a line whose intercepts are $a$ and $b$ on the coordinate axes respectively.
Let $PQ$ be a line meeting $X$ axis at $A$ and $Y$ axis at $B$. Let $OA=a$, $OB=b$. Then the coordinates of $A$ and $B$ are $(a,0)$ and $(0,b)$ respectively. Therefore, the equation of the line joining $A$ and $B$ is
A mobile phone is put to use when the battery power is 100%. The percent of battery power ‘$y$’ (in decimal) remaining after using the mobile phone for $x$ hours is assumed as $y=-0.25x+1$.
(i) Find the number of hours elapsed if the battery power is 40%.
(ii) How much time does it take so that the battery has no power?
Solution
(i) To find the time when the battery power is 40%, we have to take $y=0.40$.
A circular garden is bounded by East Avenue and Cross Road. Cross Road intersects North Street at $D$ and East Avenue at $E$. $AD$ is tangential to the circular garden at $A(3,10)$. Using the figure,
(a) Find the equation of
(i) East Avenue.
(ii) North Street.
(iii) Cross Road.
(b) Where does the Cross Road intersect?
(i) North Street (ii) East Avenue.
Fig. 5.37
Solution
(a) (i) East Avenue is the straight line joining $C(0,2)$ and $B(7,2)$. Thus the equation of East Avenue is obtained by using two-point form which is
(ii) Since the point $D$ lie vertically above $C(0,2)$. The $x$ coordinate of $D$ is 0.
Since any point on North Street has $x$ coordinate value 0.
The equation of North Street is $x=0$.
(iii) To find equation of Cross Road.
Center of circular garden $M$ is at $(7,7)$, $A$ is $(3,10)$.
We first find slope of $MA$, which we call $m_1$.
Thus
$$
m_1=\frac{10-7}{3-7}=\frac{-3}{4}.
$$
Since the Cross Road is perpendicular to $MA$, if $m_2$ is the slope of the Cross Road then, $m_1m_2=-1$ gives $\dfrac{-3}{4}m_2=-1$ so $m_2=\dfrac43$.
Now, the cross road has slope $\dfrac43$ and it passes through the point $A(3,10)$.
The equation of the Cross Road is
$$
y-10=\frac43(x-3)
$$$$
3y-30=4x-12
$$
Hence, $4x-3y+18=0$.
(b) (i) If $D$ is $(0,k)$ then $D$ is a point on the Cross Road.
Therefore, substituting $x=0$, $y=k$ in the equation of Cross Road, we get,
$$
0-3k+18=0
$$
Value of $k=6$.
Therefore, $D$ is $(0,6)$.
(ii) To find $E$, let $E$ be $(q,2)$.
Put $y=2$ in the equation of the Cross Road, we get,
Thus the Cross Road meets the North Street at $D(0,6)$ and East Avenue at $E(-3,2)$.
Progress Check
Fill the details in respective boxes.
S.No.
Equation
Slope
$x$ intercept
$y$ intercept
1
$3x-4y+2=0$
2
$y=14x$
0
3
2
$-3$
Activity 4
If line $l_1$ is perpendicular to line $l_2$ and line $l_3$ has slope 3 then
(i) find the equation of line $l_1$.
(ii) find the equation of line $l_2$.
(iii) find the equation of line $l_3$.
Fig. 5.38
Activity 5
A ladder is placed against a vertical wall with its foot touching the horizontal floor. Find the equation of the ladder under the following conditions.
No.
Condition
Picture
Equation of the ladder
(i)
The ladder is inclined at $60^\circ$ to the floor and it touches the wall at $(0,8)$
Fig. 5.39
__________
(ii)
The foot and top of the ladder are at the points $(2,4)$ and $(5,1)$
Find the equation of a straight line passing through the mid-point of a line segment joining the points $(1,-5)$, $(4,2)$ and parallel to (i) $X$ axis (ii) $Y$ axis.
The equation of a straight line is $2(x-y)+5=0$. Find its slope, inclination and intercept on the $Y$ axis.
Find the equation of a line whose inclination is $30^\circ$ and making an intercept $-3$ on the $Y$ axis.
Find the slope and $y$ intercept of $\sqrt3x+(1-\sqrt3)y=3$.
Find the value of ‘$a$’, if the line through $(-2,3)$ and $(8,5)$ is perpendicular to $y=ax+2$.
The hill in the form of a right triangle has its foot at $(19,3)$. The inclination of the hill to the ground is $45^\circ$. Find the equation of the hill joining the foot and top.
Find the equation of a line through the given pair of points
(i) $\left(2,\frac{2}{3}\right)$ and $\left(-\frac{1}{2},-2\right)$
(ii) $(2,3)$ and $(-7,-1)$
A cat is located at the point $(-6,-4)$ in $xy$ plane. A bottle of milk is kept at $(5,11)$. The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.
If the vertices of a $\triangle ABC$ are $A(6,2)$, $B(-5,-1)$ and $C(1,9)$; through the vertex $A$
(i) find the equation of median
(ii) find the equation of altitude
Find the equation of a straight line which has slope $-\frac{5}{4}$ and passing through the point $(-1,2)$.
You are downloading a song. The percent $y$ (in decimal form) of mega bytes remaining to get downloaded in $x$ seconds is given by $y=-0.1x+1$.
(i) find the total MB of the song.
(ii) after how many seconds will 75% of the song gets downloaded?
(iii) after how many seconds the song will be downloaded completely?
Find the equation of a line whose intercepts on the $x$ and $y$ axes are given below.
(i) $4,-6$
(ii) $-5,\frac{3}{4}$
Find the intercepts made by the following lines on the coordinate axes.
(i) $3x-2y-6=0$
(ii) $4x+3y+12=0$
Find the equation of a straight line
(i) passing through $(1,-4)$ and has intercepts which are in the ratio $2:5$
(ii) passing through $(-8,4)$ and making equal intercepts on the coordinate axes
The linear equation (first degree polynomial in two variables $x$ and $y$) $ax+by+c=0$ (where $a$, $b$ and $c$ are real numbers such that at least one of $a$, $b$ is non-zero) always represents a straight line. This is the general form of a straight line.
Now, let us find out the equations of a straight line in the following cases
(i) parallel to $ax+by+c=0$
(ii) perpendicular to $ax+by+c=0$
5.6.1 Equation of a line parallel to the line $ax+by+c=0$#
The equation of all lines parallel to the line $ax+by+c=0$ can be put in the form $ax+by+k=0$ for different values of $k$.
5.6.2 Equation of a line perpendicular to the line $ax+by+c=0$#
The equation of all lines perpendicular to the line $ax+by+c=0$ can be written as $bx-ay+k=0$ for different values of $k$.
Do You Know?
Two straight lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ where the coefficients are non-zero, are
(i) parallel if and only if $\frac{a_1}{a_2}=\frac{b_1}{b_2}$; That is, $a_1b_2-a_2b_1=0$
(ii) perpendicular if and only if $a_1a_2+b_1b_2=0$
Check whether the given lines are parallel or perpendicular
(i) $\frac{x}{3}+\frac{y}{4}+\frac{1}{7}=0$ and $\frac{2x}{3}+\frac{y}{2}+\frac{1}{10}=0$
(ii) $5x+23y+14=0$ and $23x-5y+9=0$
If the straight lines $12y=-(p+3)x+12$, $12x-7y=16$ are perpendicular find ‘$p$’.
Find the equation of a straight line passing through the point $P(-5,2)$ and parallel to the line joining the points $Q(3,-2)$ and $R(-5,4)$.
Find the equation of a line passing through $(6,-2)$ and perpendicular to the line joining the points $(6,7)$ and $(2,-3)$.
$A(-3,0)$, $B(10,-2)$ and $C(12,3)$ are the vertices of $\triangle ABC$. Find the equation of the altitude through $A$ and $B$.
Find the equation of the perpendicular bisector of the line joining the points $A(-4,2)$ and $B(6,-4)$.
Find the equation of a straight line through the intersection of lines $7x+3y=10$, $5x-4y=1$ and parallel to the line $13x+5y+12=0$
Find the equation of a straight line through the intersection of lines $5x-6y=2$, $3x+2y=10$ and perpendicular to the line $4x-7y+13=0$
Find the equation of a straight line joining the point of intersection of $3x+y+2=0$ and $x-2y-4=0$ to the point of intersection of $7x-3y=-12$ and $2y=x+3$
Find the equation of a straight line through the point of intersection of the lines $8x+3y=18$, $4x+5y=9$ and bisecting the line segment joining the points $(5,-4)$ and $(-7,6)$.
The area of triangle formed by the points $(-5,0)$, $(0,-5)$ and $(5,0)$ is
(A) 0 sq.units (B) 25 sq.units (C) 5 sq.units (D) none of these
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the $Y$ axis. The path travelled by the man is
(A) $x=10$ (B) $y=10$ (C) $x=0$ (D) $y=0$
The straight line given by the equation $x=11$ is
(A) parallel to $X$ axis (B) parallel to $Y$ axis (C) passing through the origin (D) passing through the point $(0,11)$
If $(5,7)$, $(3,p)$ and $(6,6)$ are collinear, then the value of $p$ is
(A) 3 (B) 6 (C) 9 (D) 12
The point of intersection of $3x-y=4$ and $x+y=8$ is
(A) $(5,3)$ (B) $(2,4)$ (C) $(3,5)$ (D) $(4,4)$
The slope of the line joining $(12,3)$, $(4,a)$ is $\frac{1}{8}$. The value of ‘$a$’ is
(A) 1 (B) 4 (C) $-5$ (D) 2
The slope of the line which is perpendicular to a line joining the points $(0,0)$ and $(-8,8)$ is
(A) $-1$ (B) 1 (C) $\frac{1}{3}$ (D) $-8$
If slope of the line $PQ$ is $\frac{1}{\sqrt{3}}$ then slope of the perpendicular bisector of $PQ$ is
(A) $l_1$ and $l_2$ are perpendicular (B) $l_1$ and $l_4$ are parallel (C) $l_2$ and $l_4$ are perpendicular (D) $l_2$ and $l_3$ are parallel
A straight line has equation $8y=4x+21$. Which of the following is true
(A) The slope is 0.5 and the $y$ intercept is 2.6 (B) The slope is 5 and the $y$ intercept is 1.6 (C) The slope is 0.5 and the $y$ intercept is 1.6 (D) The slope is 5 and the $y$ intercept is 2.6
When proving that a quadrilateral is a trapezium, it is necessary to show
(A) Two sides are parallel. (B) Two parallel and two non-parallel sides. (C) Opposite sides are parallel. (D) All sides are of equal length.
When proving that a quadrilateral is a parallelogram by using slopes you must find
(A) The slopes of two sides (B) The slopes of two pair of opposite sides (C) The lengths of all sides (D) Both the lengths and slopes of two sides
$(2,1)$ is the point of intersection of two lines.
PQRS is a rectangle formed by joining the points $P(-1,-1)$, $Q(-1,4)$, $R(5,4)$ and $S(5,-1)$. $A$, $B$, $C$ and $D$ are the mid-points of $PQ$, $QR$, $RS$ and $SP$ respectively. Is the quadrilateral $ABCD$ a square, a rectangle or a rhombus? Justify your answer.
The area of a triangle is 5 sq.units. Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex is $(x,y)$ where $y=x+3$. Find the coordinates of the third vertex.
Find the area of a triangle formed by the lines $3x+y-2=0$, $5x+2y-3=0$ and $2x-y-3=0$
If vertices of a quadrilateral are at $A(-5,7)$, $B(-4,k)$, $C(-1,-6)$ and $D(4,5)$ and its area is 72 sq.units. Find the value of $k$.
Without using distance formula, show that the points $(-2,-1)$, $(4,0)$, $(3,3)$ and $(-3,2)$ are vertices of a parallelogram.
Find the equations of the lines, whose sum and product of intercepts are 1 and $-6$ respectively.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?
Find the image of the point $(3,8)$ with respect to the line $x+3y=7$ assuming the line to be a plane mirror.
Find the equation of a line passing through the point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.
A person standing at a junction (crossing) of two straight paths represented by the equations $2x-3y+4=0$ and $3x+4y-5=0$ seek to reach the path whose equation is $6x-7y+8=0$ in the least time. Find the equation of the path that he should follow.
Step 1: Open the Browser type the URL Link given below (or) Scan the QR Code. GeoGebra work book named “Co-Ordinate Geometry” will open. In the left side of the work book there are many activity related to mensuration chapter. Select the work sheet “Area of a Quadrilateral”
Step 2: In the given worksheet you can change the Question by clicking on “New Problem”. Move the slider to see the steps. Work out each problem and verify your answer.
Step 1: Open the Browser type the URL Link given below (or) Scan the QR Code. GeoGebra work book named “Co-Ordinate Geometry” will open. In the left side of the work book there are many activity related to mensuration chapter. Select the work sheet “Slope_Equation of a Straight Line”
Step 2: In the given worksheet you can change the Line by Dragging the points A and B on graph. Click on the Check boxes on Left Hand Side to see various forms of same straight line.
ICT 5.2 — Step 1, Step 2 and Expected results
You can repeat the same steps for other activities
