Coordinate Geometry: Question 7

Worked solution

Part (a): Equation of $l_1$

Gradient. For $A(-2, 5)$ and $B(4, -1)$:

$m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1.$

Equation. Using point $A(-2, 5)$ in $y - y_1 = m(x - x_1)$:

$y - 5 = -1(x - (-2)) = -1(x + 2) = -x - 2$

$y = -x - 2 + 5 = -x + 3.$

Check with $B(4, -1)$: $y = -(4) + 3 = -1. \checkmark$

So $l_1$ has equation $\boxed{y = -x + 3}$.

Part (b): Equation of $l_2$

Since $l_2$ is parallel to $l_1$, it has the same gradient, $m = -1$.

Using the point $C(5, 4)$:

$y - 4 = -1(x - 5) = -x + 5$

$y = -x + 5 + 4 = -x + 9.$

Check with $C(5, 4)$: $y = -(5) + 9 = 4. \checkmark$

So $l_2$ has equation $\boxed{y = -x + 9}$.

Part (c): Intersection of $l_1$ and $l_3$

At the point of intersection, the $y$-values of $l_1$ and $l_3$ are equal:

$-x + 3 = 2x - 3$

$3 + 3 = 2x + x$

$6 = 3x \implies x = 2.$

Substituting $x = 2$ back into $l_1$‘s equation:

$y = -(2) + 3 = 1.$

Check with $l_3$: $y = 2(2) - 3 = 4 - 3 = 1. \checkmark$ Both equations agree, confirming the intersection point.

So $l_1$ and $l_3$ intersect at $\boxed{(2, 1)}$.

Final answers

• (a) $l_1$: $y = -x + 3$
• (b) $l_2$: $y = -x + 9$
• (c) $l_1$ and $l_3$ intersect at $(2, 1)$