Question

find the equation if the line which is perpendicular to the given line 3x-4y=1 and passes through (3,4)

Answer 1

Given line =3x-4y=1
3x-1=4y
On comparing with y=mx+c
Slope =3/4
m1×m2=-1
m2=-4/3
It passes through (3,4)
y-y1=m(x-x1)
y-4=-4/3(x-3)
3y-12=-4x+12
4x+3y-24=0

Answer 2

GIVEN :-

• Equation of a line (l₁) is 3x - 4y = 1

• A line (l₂) passes through point ( 3 , 4 )

TO FIND :-

• Equation of line (l₂) which is perpendicular to Line (l₂)

SOLUTION :-

When the given two lines are perpendicular to each other,Then the relation between their Slopes m₁ and m₂ is given by ,

$\large {\underline {\bold {\boxed {\bigstar {\red {\sf{ \: m_1 \times m_2 = - 1}}}}}}}$

Where ,

• m₁ is the slope of the line l₁

• m₂ is the slope of the line l₂

We have ,

• Equation of line (l₂) is 3x - 4y = 1

$\implies \sf \: -4y = 1 - 3x \\ \\ \implies \sf \: y = -\frac{1}{4} + \frac{3x}{4} \\ \\ \implies \underline {\bold {\boxed {\pink{\sf{ \: m_2 = + \frac{3}{4} }}}}}$

From the above relation,

$\implies \sf \: m_1 \times - \dfrac{ 3}{4} = - 1 \\ \\ \implies \sf \: m_1 = \frac{ - 1}{ \frac{ 3}{4} } \\ \\ \implies \sf \: m_1 = \frac{ - 1(4)}{ 3} \\ \\ \implies {\underline {\bold {\boxed {\pink{ \sf m_1 = \frac{-4}{3} }}}}}$

The line which passes through the point (x₁ , y₁ ) and having slope 'm' Then the equation of the lines is given by,

$\large {\underline {\bold {\boxed {\bigstar {\red {\sf{ \: (y - y_1) = m(x - x_1)}}}}}}}$

We are given that ,The line is passing through the point (3,4) and slope m₁ = -4/3 . By comparing we get ,

• x₁ = 3 , y₁ = 4
• m = -4/3

$\implies \sf \: (y - 4) = \dfrac{-4}{3} (x - 3) \\ \\ \implies \sf \: 3(y - 4) = -4(x - 3) \\ \\ \implies \sf \: 3y - 12 = -4x +12 \\ \\ \implies \sf 3y - 12 + 4x -12 = 0 \\ \\ \implies {\underline {\bold {\boxed {\blue {\sf {3y + 4x = 24}}}}}}$

∴ The equation of the line perpendicular to line 3x - 4y = 1 and passing through the point (3,4) is 3y + 4x = 24