The equation of a straight line passing through (1, 2) and perpendicular to the lin3x + 4y = 7 is:
4x - 3y + 2 = 0
3x + 4y = 7
4x - 3y + 7 = 0
None of these
Solution
Given line, 3x + 4y = 7
Or y $=~-\frac{3x}{4}~+\frac{7}{3}$
Comparing this equation with $y~=~mx~+~c,~m~=~-\frac{3}{4}$
The slope of the line perpendicular to this line,
$m\_1~~=~-\frac{1}{m}~=\frac{4}{3}$
The equation of a line passing through (${{x}_{1}},~{{x}_{2}})~$and a slope
of m is given by the formula $(y~~{{y}_{1}})~=~m(x~~{{x}_{1}}).$
The equation of the line passing through (1, 2) and as slope of 4/3 is:
$\left( y~-~2 \right)=\frac{4}{3}\times ~(x~-~1)$
So, 4x - 3y + 2 = 0