Question

A(6,4), B(8,4), C(10,6) and D(2,8) are the vertices of quadrilateral ABCD. Then the point of intersection of its diagonals is

• A. $(\frac{5}{7},\frac{2}{7})$
• B. $(\frac{15}{7},\frac{19}{7})$
• C. $(\frac{50}{7},\frac{32}{7})$
• D. $(\frac{15}{7},\frac{32}{7})$

Answer 1

The correct option is C

$(\frac{50}{7},\frac{32}{7})$

We know that the slope(m) of the line formed by joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is given by,

$m=\frac{y_2-y_1}{x_2-x_1}$

Now, the slope of the diagonal AC

$=\frac{6-4}{10-6}$

$=\frac{1}{2}$

And the slope of the diagonal BD is

$=\frac{8-4}{2-8}$

$=-\frac{2}{3}$

Let $P(x,y)$ be the point of intersection of these diagonals.

Then $P(x,y)$ lies on both AC and BD.

Therefore $\frac{y-6}{x-10}=\frac{1}{2}$ and $\frac{y-8}{x-2}=-\frac{2}{3}$

$⟹2y-12=x-10$ and $3y-24=-2x+4$

$⟹2y-x=2$ and $3y+2x=28$

Solving the above two equations, we get, $x=\frac{50}{7}$ and $y=\frac{32}{7}$

Hence the point of intersection P of the diagonals AC and BD is $(\frac{50}{7},\frac{32}{7})$