Question

In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively. 

1. Write the co-ordinates of A. 
2. Find the length of AB and AC.
3. Find the radio in which Q divides AC. 
4. Find the equation of the line AC.

Answer 1

i. The line intresects the x-axis where y = 0

Hence, the co-ordinates of A are (4, 0)

ii. Length of AB = ${\displaystyle \sqrt{{(-2-4)}^2+{(3-0)}^2}}$

= ${\displaystyle \sqrt{36+9}}$

= ${\displaystyle \sqrt{45}}$

= ${\displaystyle 3\sqrt{5}}$ units

Length of AC = ${\displaystyle \sqrt{{(-2-4)}^2+{(-4-0)}^2}}$

= ${\displaystyle \sqrt{36+16}}$

= ${\displaystyle \sqrt{52}}$

= ${\displaystyle 2\sqrt{13}}$ units

iii. Let K be the required ratio which divides the line segment joining the co-ordinates A(4, 0) and O(–2, −4)

Let the co-ordinates of Q be x and y 

∴ ${\displaystyle x=\frac{k(-2)+1\left(4\right)}{k+1}}$ and ${\displaystyle y=\frac{k(-4)+0}{k+1}}$

Q lies on the y-axis where x = 0,

${\displaystyle \Rightarrow \frac{-2k+4}{k+1}=0}$

${\displaystyle \Rightarrow }$ –2k + 4 = 0

${\displaystyle \Rightarrow }$ 2k = 4

${\displaystyle \Rightarrow k=\frac{4}{2}=\frac{2}{1}}$

Thus the required ratio is 2 : 1

iv. Slope of line AC = m = ${\displaystyle \frac{-4-0}{-2-4}=\frac{-4}{-6}=\frac{2}{3}}$

Thus, the equation of the line AC is given by 

${\displaystyle y-0=\frac{2}{3}(x-4)}$

i.e 3y = 2x – 8

i.e 2x – 3y = 8