Question 12 Marks
Find the co-ordinates of the centroid of a triangle ABC whose vertices are: $A(-1, 3), B(1, -1)$ and $C(5, 1)$
AnswerCo- ordinates of the centroid of triangle ABC are
$\left(\frac{-1+1+5}{3}, \frac{3-1+1}{3}\right)$
$=\left(\frac{5}{3}, 1\right)$
View full question & answer→Question 22 Marks
The mid-point of the segment AB, as shown in diagram, is C$(4, -3).$ Write down the co-ordinates of A and B.
AnswerSince, point A lies on x-axis, let the co-ordinates of point A be $(x, 0).$
Since, point B lies on y-axis, let the co-ordinates of point B be $(0, y).$
Given, mid-point of AB is C $(4, −3).$
$\therefore(4,-3)=\left(\frac{x+0}{2}, \frac{0+y}{2}\right) $
$ \Rightarrow(4-3)=\left(\frac{x}{2}, \frac{y}{2}\right) $
$\Rightarrow 4=\frac{x}{2} \text { and }-3=\frac{y}{2}$
$ \Rightarrow x=8 \text { and } y=-6$
Thus, the co-ordinates of point A are $(8, 0)$ and the co-ordinates of point B are $(0, −6).$
View full question & answer→Question 32 Marks
Find the image of the point A$(5,3)$ under reflection in the point P$(-1,3).$
AnswerLet the image be $B(x, y)$.
Since A is reflected in P,P is the mid-point of AB
Using mid-point formula, we get
$\Rightarrow P(-1,3)=\left(\frac{x+5}{2}, \frac{y-3}{2}\right)$
$\Rightarrow \frac{x+5}{2}=-1 \text { and } \frac{y-3}{2}=3 $
$ \Rightarrow x+5=-2 \text { and } y-3=6$
$ \Rightarrow x=-7 \text { and } y=9$
So, the image of A in P is $B(-7,9)$
View full question & answer→Question 42 Marks
One end of the diameter of a circle is $(-2, 5).$ Find the co-ordinates of the other end of it, of the centre of the circle is $(2, -1)$
AnswerWe know that the centre is the mid-point of diameter.
Let the required co-ordinates of the other end of mid-point be $(x, y).$
$\therefore(2,-1)=\left(\frac{-2+x}{2}, \frac{5+y}{2}\right)$
$\Rightarrow 2=\frac{-2+x}{2} \text { and }-1=\frac{5+y}{2} $
$ \Rightarrow 4=-2+x \text { and }-2=5+y $
$ \Rightarrow 6=x \text { and }-7=y$
Thus, the required co-ordinates are $(6, −7).$
View full question & answer→Question 52 Marks
In the given figure, $P (4, 2)$ is mid-point of line segment $AB.$ Find the co-ordinates of $A$ and $B.$
AnswerPoint $A$ lies on x-axis, so let its co-ordinates be $(x, 0).$
Point $B$ lies on y-axis, so let its co-ordinates be $(0, y).$
$P\ (4, 2)$ is mid-point of line segment $AB.$
$\therefore(4,2)=\left(\frac{x+0}{2}, \frac{0+y}{2}\right)$
$\Rightarrow 4=\frac{x}{2} \text { and } 2=\frac{y}{2} $
$ \Rightarrow 8=x \text { and } 4=y$
Hence, the co-ordinates of points A and B are $(8, 0)$ and $(0, 4)$ respectively.
View full question & answer→Question 62 Marks
Given M is the mid point of AB, find the co-ordinates of: B; if $A = (3, -1)$ and $M = (-1, 3)$
AnswerLet the co-ordinates of B be (x, y).
$\therefore(-1,3)=\left(\frac{3+x}{2}, \frac{-1+y}{2}\right)$
$\Rightarrow-1=\frac{3+x}{2}$ and $3=\frac{-1+y}{2}$
$\Rightarrow-2=3+x \text { and } 6=-1+y $
$ \Rightarrow x=-5 \text { and } y=7$
Hence, the co-ordinates of B are $(-5, 7).$
View full question & answer→Question 72 Marks
Given $M$ is the mid point of $AB,$ find the co-ordinates of $: A;$ if $M = (1, 7)$ and $B = (-5, 10)$
AnswerLet the co-ordinates of $A$ be $(x, y).$
$\therefore(1,7)=\left(\frac{x-5}{2}, \frac{y+10}{2}\right)$
$\Rightarrow 1=\frac{x-5}{2} \text { and } 7=\frac{y+10}{2}$
$\Rightarrow 2=(x-5) \text { and } 14=y+10$
$\Rightarrow x=7 \text { and } y=4$
Hence, the co-ordinates of A are $(7, 4).$
View full question & answer→Question 82 Marks
Points A and B have co-ordinates $(3, 5)$ and $(x, y)$ respectively. The mid point of AB is $(2, 3).$ Find the values of x and y.
AnswerMid-point of $AB = (2, 3)$
$\therefore\left(\frac{3+x}{2}, \frac{5+y}{2}\right)=(2,3)$
$\Rightarrow \frac{3+x}{2}=2 \text { and } \frac{5+y}{2}=3 $
$ \Rightarrow 3+x=4 \text { and } 5+y=6 $
$ \Rightarrow x=1 \text { and } y=1$
View full question & answer→Question 92 Marks
$A (5, x), B (-4, 3)$ and $C\ (y, -2)$ are the vertices of the $\triangle ABC$ whose centroid is the origin. Calculate the values of $x$ and $y.$
AnswerGiven, centroid of $\triangle ABC$ is the origin.
$\therefore(0,0)=\left(\frac{5-4+y}{3}, \frac{x+3-2}{3}\right)$
$(0,0)=\left(\frac{1+y}{3}, \frac{x+1}{3}\right)$
$0=\frac{1+y}{3} \text { and } 0=\frac{x+1}{3}$
$y=-1 \text { and } x=-1$
View full question & answer→Question 102 Marks
The co-ordinates of the centroid of a triangle $PQR$ are $(2, -5).$ If $Q = (-6, 5)$ and $R = (11, 8);$ calculate the co-ordinates of vertex $P.$
AnswerLet $G$ be the centroid of $DPQR$ whose coordinates are $(2, −5)$ and let $(x, y)$ be the coordinates of vertex $P.$ Coordinates of $G$ are,
$G(2,-5)=G\left(\frac{x-6+11}{3}, \frac{y+5+8}{3}\right)$
$2=\frac{x+5}{3},-5=\frac{y+13}{3}$
$6=x+5,-15=y+13$
$x=1, y=-28$ Coordinates of vertex $P$ are $(1, −28)$
View full question & answer→Question 112 Marks
Calculate the co-ordinates of the centroid of the triangle ABC, if $A = (7, -2), B = (0, 1)$ and $C =(-1, 4).$
AnswerCo-ordinates of the centroid of triangle $ABC$ are
$\left(\frac{7+0-1}{3}, \frac{-2+1+4}{3}\right)$
$=\left(\frac{6}{3}, \frac{3}{3}\right)$
$=(2,1)$
View full question & answer→Question 122 Marks
Points P (a, −4), Q (−2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of a and b.
AnswerGiven, PR = 2QR
Now, Q lies between P and R, so, PR = PQ + QR
∴ PQ + QR = 2QR
⇒ PQ=QR
⇒ Q is the mid-point of PR.
$\therefore(-2, b)=\left(\frac{a+0}{2}, \frac{-4+2}{2}\right)$
$(-2, b)=\left(\frac{a}{2},-1\right)$
$\Rightarrow a=-4$ and $b=-1$
View full question & answer→Question 132 Marks
Points $A (-5, x), B (y, 7)$ and $C (1, -3)$ are collinear (i.e. lie on the same straight line) such that $AB = BC.$ Calculate the values of $x$ and $y.$
AnswerGiven, $AB = BC,$ i.e., $B$ is the mid-point of $AC.$
$\therefore (y, 7)=\left(\frac{-5+1}{2},(x-3), 2\right)$
$(y, 7)=\left(-2, \frac{x-3}{2}\right)$
$\Rightarrow y=-2 \text { and } 7=\frac{x-3}{2}$
$\Rightarrow y=-2 \text { and } x=17$
View full question & answer→Question 142 Marks
Find the mid – point of the line segment joining the point: (5, -3) and (-1, 7)
AnswerA (5, -3) and B (-1, 7)
Mid-point of AB $=\left(\frac{5-1}{2}, \frac{-3+7}{2}\right)$
Mid-point of $A B=(2,2)$
View full question & answer→Question 152 Marks
Find the mid – point of the line segment joining the point: (-6, 7) and (3, 5)
AnswerA (-6, 7) and B (3, 5)
Mid-point of AB $=\left(\frac{-6+3}{2}, \frac{7+5}{2}\right)$
$=\left(-\frac{3}{2}, 6\right)$
View full question & answer→Question 162 Marks
A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the coordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
AnswerCo-ordinates of the mid-point of AC are
$\left(\frac{2-4}{2}, \frac{5+3}{2}\right)=\left(-\frac{2}{2}, \frac{8}{2}\right)=(-1,4)$
Co-ordinates of the mid-point of BD are
$\left(\frac{1-3}{2}, \frac{0+8}{2}\right)=\left(-\frac{2}{2}, \frac{8}{2}\right)=(-1,4)$
Since, mid-point of AC = mid-point of BD
Hence, ABCD is a parallelogram.
View full question & answer→Question 172 Marks
The line joining the points $A (-3, -10)$ and $B (-2, 6)$ is divided by the point $P$ such that $\frac{P B}{A B}=\frac{1}{5}$ Find the co-ordinates of $P$.
AnswerLet the co-ordinates of point $P$ are $(x, y).$
Given: $PB : AB = 1 : 5$
$\therefore PB : PA = 1 : 4$
Coordinates of $P$ are
$(x, y)=\left(\frac{4 \times(-2)+1 \times(-3)}{5}, \frac{4 \times 6+1 \times(-10)}{5}\right)$
$=\left(-\frac{11}{5}, \frac{14}{5}\right)$ View full question & answer→Question 182 Marks
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates of the point of intersection.
Answer
$0=\frac{3 k-4}{k+1}$
$3 k=4$
$k=\frac{4}{3}......(1)$
$y=\frac{0+7}{k+1}$
$y=\frac{7}{\frac{4}{3}+1} \quad$ (from eq. 1 )
$y=3$
Hence, the required is 4:3 and the required point is S(0, 3) View full question & answer→Question 192 Marks
In what ratio is the line joining $(2, -3)$ and $(5, 6)$ divided by the x-axis.
AnswerLet the line joining points A $(2, −3)$ and B $(5, 6)$ be divided by point P $(x, 0)$ in the ratio $k : 1.$
$y=\frac{k y_2+y_1}{k+1} $
$ 0=\frac{k \times 6+1 \times(-3)}{k+1}$
$ 0=6 k-3 $
$ k=\frac{1}{2}$
Thus, the required ratio is $1: 2.$
View full question & answer→Question 202 Marks
Calculate the ratio in which the line joining $A(6, 5)$ and $B(4, -3)$ is divided by the line $y = 2.$
AnswerThe co-ordinates of every point on the line $y = 2$ will be of the type $(x, 2).$
Using section formula, we have:
$y=\frac{m_1 \times(-3)+m_2 \times 5}{m_1+m_2} $
$ 2=\frac{-3 m_1+5 m_2}{m_1+m_2}$
$2 m_1+2 m_2=-3 m_1+5 m_2$
$5 m_1 =3 m_2 $
$ \frac{m_1}{m_2} =\frac{3}{5}$
Thus, the required ratio is $3 : 5.$
View full question & answer→Question 212 Marks
P is a point on the line joining $A(4, 3)$ and $B(-2, 6)$ such that $5AP = 2BP.$ Find the co-ordinates of P.
Answer$5AP = 2BP$
$\frac{A P}{B P}=\frac{2}{5}$
The co-ordinates of the point P are
$\left(\frac{2 \times(-2)+5 \times 4}{2+5}, \frac{2 \times 6+5 \times 3}{2+5}\right) $
$ \left(\frac{16}{7}, \frac{27}{7}\right)$
View full question & answer→