Show that the line $3 x + 4 y + 7 = 0$ and $28 x - 21 y + 50 = 0$ are perpendicular to each other.

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Solution

We know that the equation of a line is $y = m x + c$ where $m$ is the slope of the line and $c$ is the y-intercept.

We solve the first line $3 x + 4 y + 7 = 0$ for $y$ as follows:

$3 x + 4 y + 7 = 0 \Rightarrow 4 y = - 3 x - 7 \Rightarrow y = \frac{- 3 x - 7}{4} \Rightarrow y = - \frac{3 x}{4} - \frac{7}{4} . . . . . . (1)$

Comparing the equation of line $y = m x + c$ with equation 1 we have the slope of first line that is $m_{1} = - \frac{3}{4}$ .

Similarly, we solve the second line $28 x - 21 y + 50 = 0$ for $y$ as follows:

$28 x - 21 y + 50 = 0 \Rightarrow 21 y = 28 x + 50 \Rightarrow y = \frac{28 x + 50}{21} \Rightarrow y = \frac{28 x}{21} + \frac{50}{21} \Rightarrow y = \frac{4}{3} x + \frac{50}{21} . . . . . . (2)$

Comparing the equation of line $y = m x + c$ with equation 2 we have the slope of second line that is $m_{2} = \frac{4}{3}$ .

Now, consider the product of the slopes $m_{1}$ and $m_{2}$ that is:

$m_{1} \times m_{2} = - \frac{3}{4} \times \frac{4}{3} = - 1$ and

Since $m_{1} \times m_{2} = - 1$ and we know that if the slope of the two lines have the relation $m_{1} \times m_{2} = - 1$ , then the lines are perpendicular to each other.

Hence, the lines $3 x + 4 y + 7 = 0$ and $28 x - 21 y + 50 = 0$ are perpendicular to each other.

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Show that the line 3x+4y+7=0 and 28x−21y+50=0 are perpendicular to each other.

Show that the line $3 x + 4 y + 7 = 0$ and $28 x - 21 y + 50 = 0$ are perpendicular to each other.