Question

In the given figure, ABC is a triangle and BC is parallel to $$y$$-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 ratio in which Q divides AC.
4. Find the equation of the line AC.

Answer 1

i. In the given figure, point $$A$$ lies on the positive $$x$$-axis and $$\triangle ABC$$ is symmetrical about the $$x$$-axis or from the grid coordinates $$A$$ is located at $$(4, 0)$$.

Thus, the coordinates of $$A$$ are $$(4, 0)$$.

ii. Coordinates of the vertices are $$A(4, 0)$$, $$B(-2, 3)$$ and $$C(-2, -4)$$.

Length of $$AB$$: 

$$AB = \sqrt{(-2 - 4)^2 + (3 - 0)^2}$$ 

$$= \sqrt{(-6)^2 + 3^2}$$ 

$$= \sqrt{36 + 9}$$ 

$$= \sqrt{45}$$ 

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

Length of $$AC$$:

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

$$= \sqrt{(-6)^2 + (-4)^2}$$

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

$$= \sqrt{52}$$

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

iii. Point $$Q$$ is the intersection of $$AC$$ with the $$y$$-axis, so its $$x$$-coordinate is $$0$$. 

Let $$Q$$ divide the segment $$AC$$ in the ratio $$m_1 : m_2$$, where $$A(4, 0)$$ and $$C(-2, -4)$$. 

Using the section formula for the $$x$$-coordinate: 

$$x_Q = \frac{m_1(-2) + m_2(4)}{m_1 + m_2} = 0$$ 

$$-2m_1 + 4m_2 = 0$$ 

$$2m_1 = 4m_2$$ 

$$\frac{m_1}{m_2} = \frac{4}{2} = \frac{2}{1}$$ 

Thus, $$Q$$ divides $$AC$$ in the ratio $$2 : 1$$.

iv. Equation of the line $$AC$$ passing through $$A(4, 0)$$ and $$C(-2, -4)$$:

Slope of $$AC$$: 

$$m = \frac{-4 - 0}{-2 - 4}$$ 

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

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

Using the point-slope form with $$A(4, 0)$$: 

$$y - 0 = \frac{2}{3}(x - 4)$$ 

$$3y = 2(x - 4)$$

$$3y = 2x - 8$$

$$2x - 3y - 8 = 0$$