First slide

Equation of line in Straight lines

Question

The straight line 3x + 4y - 12=0 meets the coordinate axes at A and B. An equilateral triangle ABC is constructed. The possible coordinates of vertex C are

Moderate

A

$(2 (1 - \frac{3 \sqrt{3}}{4}), \frac{3}{2} (1 - \frac{4}{\sqrt{3}}))$
B

$(- 2 (1 + \sqrt{3}), \frac{3}{2} (1 - \sqrt{3}))$
C

$(2 (1 + \sqrt{3}), \frac{3}{2} (1 + \sqrt{3}))$
D

$(2 (1 + \frac{3 \sqrt{3}}{4}), \frac{3}{2} (1 + \frac{4}{\sqrt{3}}))$

Solution

$\begin{aligned} AB = 5, D = (2, \frac{3}{2}) \\ CD = 5 \times \frac{\sqrt{3}}{2} = \frac{5 \sqrt{3}}{2} \end{aligned}$
Slope of $AB = - \frac{3}{4}$
Slope of $CD = \frac{4}{3}$
lf C = (h, k), then
$\begin{aligned} \frac{h - 2}{3 / 5} = \frac{k - 3 / 2}{4 / 5} = \pm \frac{5 \sqrt{3}}{2} \\ or & h = 2 (1 - \frac{3 \sqrt{3}}{4}), k = \frac{3}{2} (1 - \frac{4}{\sqrt{3}}) \\ or & h = 2 (1 + \frac{3 \sqrt{3}}{4}), k = \frac{3}{2} (1 + \frac{4}{\sqrt{3}}) \end{aligned}$

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