The equation of a line is $3 x + 4 y - 7 = 0$ . Find
(i) the slope of the line
(ii) the equation of a line perpendicular to the given line ad passing through the intersection of the lines $x - y + 2 = 0$ and $3 x + y - 10 = 0$

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Solution

Slope of $3 x + 4 y - 7 = 0$

$m_{1} = - \frac{y}{x} = - \frac{y}{3}$

Since desired line passes through $x - y + z = 0$ and $3 x + y - 10 = 0$

On solving, $x = 2$ ; $y = 0$

Equation of line which is perpendicular to $3 x + 4 y - 7 = 0$
$m_{2} (x - 2) = (y - 4)$ , where $m_{2}$ is slope

$m_{2} = \frac{3}{4}$

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

$4 y - 16 = 3 x - 6$

$3 x - 4 y + 10 = 0$

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Similar questions

Equation of the straight line passing through the point of intersection of the lines $3 x + 4 y = 7, x - y + 2 = 0$ and having slope $3$ is

x + 2 y + 3 = 0

3 x + 4 y + 7 = 0

y - x = 8

3 x + 4 y = 7

x - y + 2 = 0

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