waecmaths question:
Find the equation of a straight line passing through the point (1, −5) and having gradient $\tfrac{3}{4}$
Option A:
$3x+4y-23=0$
Option B:
$3x+4y+23=0$
Option C:
$3x-4y+23=0$
Option D:
$3x-4y-23=0$
waecmaths solution:
$\begin{align} & \text{Using one point form equation of a line} \\ & y-{{y}_{1}}=m(x-{{x}_{1}}) \\ & ({{x}_{1}},{{y}_{1}})=(1,-5),\text{ }m=\tfrac{3}{4} \\ & y-(-5)=\tfrac{3}{4}(x-1) \\ & y+5=\tfrac{3}{4}(x-1) \\ & 4(y+5)=3(x-1) \\ & 4y+20=3x-3 \\ & 3x-4y-23=0 \\\end{align}$
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