Question

i)The line joining $A(2,3)$ and $B(6,-5)$ meets the x-axis at $P$. Write down the y-coordinate of $P$. ii) Hence find the ratio $AP:PB$.
iii)In the given figure $P(2,3)$ is the mid-point of the line $AB$. Write down the coordinates of $A$ and $B$.

• A. $i)3.5$ $ii)5:3$, $A=(4,0)$ $and$ $B=(0,6)$
• B. $i)0$ $ii)5:3$, $A=(0,4)$ $and$ $B=(0,6)$
• C. $i)3.5$, $ii)3:5$, $A=(4,0)$ $and$ $B=(6,0)$
• D. $i)0$, $ii)3:5$, $A=(4,0)$ $and$ $B=(0,6)$

Answer 1

The correct option is D $i)0$, $ii)3:5$, $A=(4,0)$ $and$ $B=(0,6)$

(i) Since the line joining A(2,3) and B(6,-5) meets the x-axis at P, then the y-coordinate of P is zero(0).

(ii) Let the coordinate of point P is (x,0), also the point P lies between A and B.

Let the ratio of AP:PB be k:1

By section formula,

$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}\Rightarrow x=\frac{1.6+k.2}{k+1}=\frac{6+2k}{k+1}and,y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\Rightarrow 0=\frac{k.(-5)+1.3}{k+1}\Rightarrow k=\frac{3}{5}$

therefore ratio of AP:PB=3:5

(iii) since P(2,3) is the midpoint of A(x',0) and B(0,y') (from the figure)

$\therefore x=\frac{x_1+x_2}{2}\Rightarrow x^{\prime }=4and,y=\frac{y_1+y_2}{2}\Rightarrow y^{\prime }=6$

therefore coordinate of A and B is (4,0) and (0,6) respectively.