Question

Find the equation to the straight line passing through the point of intersection of the lines 5x–6y –1 = 0 and 3x + 2y +5 =0 and perpendicular to the line 3x–5y+11 = 0

Answer 1

Answer on Question #62348 – Math – Analytic Geometry

Question

Find the equation to the straight line passing through the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$ and perpendicular to the line $3x - 5y + 11 = 0$.

Solution

Let's find the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$:

$$\left\{ \begin{array}{l} 5x - 6y - 1 = 0, \\ 3x + 2y + 5 = 0. \end{array} \right.$$

Multiply the second equation by 3:

$$\left\{ \begin{array}{l} 5x - 6y - 1 = 0, \\ 9x + 6y + 15 = 0. \end{array} \right.$$

Adding two equations

$$14x + 14 = 0.$$

Then

$$14x = -14.$$

Hence

$$x = -1.$$

Substituting $x = -1$ into the equation (1):

$$\begin{array}{l} 5 \cdot (-1) - 6y - 1 = 0; \\ -6 - 6y = 0. \\ \end{array}$$

Then

$$6y = -6.$$

Hence

$$y = -1.$$

Hence the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$ is $(-1; -1)$.

The equation of the line in the form $y = kx + b$ is called the slope-intercept form, because $k$ is the slope and $b$ gives the $y$-intercept.

Find the slope of the line $3x - 5y + 11 = 0$.

Dividing $5y = 3x + 11$ by 5:

$$y = \frac{3}{5}x + \frac{11}{5}.$$

Then the slope is

$$k = \frac{3}{5}.$$

If two lines are perpendicular, their slopes $k$ and $k_1$ are negative reciprocals.

It means

$$k \cdot k_1 = -1.$$

We get

$$k_1 = -\frac{1}{k} = -\frac{1}{\frac{3}{5}} = -\frac{5}{3}.$$

The general equation of the line through the point $(x_1; y_1)$ with the slope $k_1$ is given by

$$y - y _ {1} = k _ {1} \left(x - x _ {1}\right).$$

Let's find the equation of the line that passes through the point $(-1; -1)$ with the slope $k_{1} = -\frac{5}{3}$ :

$$y + 1 = - \frac {5}{3} (x + 1).$$

Multiplying by 3:

$$3 y + 3 = - 5 (x + 1);$$ $$3 y + 3 = - 5 x - 5.$$

The equation to the straight line is

$$5 x + 3 y + 8 = 0.$$

Answer: $5x + 3y + 8 = 0$

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