Question

Write the general equation of the line satisfying the given conditions:
1. passing trough (1, 0) and parallel to the line through (1/2, 3) and (0,0)
2. passing through (2, -4) and parallel to the line 3x - 7y - 11 = 0
3. passing through (-1/3, 1/2) and perpendicular to the line through (4, 1) and (-2, 7)
4. passing through (-1/2, 2) and perpendicular to the line with slope -5/2
5. passing through the point of intersection of lines 5x - 2y - 12 = 0 and x + 3y + 1 = 0 and perpendicular to the line 7x + 4y - 10 = 0.
6.The base of a triangle is the line segment from (3,0) to (2,-3). If the area of the triangle is 7, find the general equation of the locus of the third vertex. (Two solutions).

Find the distance between the given lines:
a. 3x - y = 5 and 3x - y = 25
b. 4x - 6y = 9 and 2x - 3y = 6

Answer 1

Answer on question №57671, Math / Analytic Geometry

Task:

Write the general equation of the line satisfying the given conditions:

1. passing through (1, 0) and parallel to the line through (1/2, 3) and (0,0)

2. passing through (2, -4) and parallel to the line $3x - 7y - 11 = 0$

3. passing through $(-1/3, 1/2)$ and perpendicular to the line through (4, 1) and (-2, 7)

4. passing through $(-1/2, 2)$ and perpendicular to the line with slope $-5/2$

5. passing through the point of intersection of lines $5x - 2y - 12 = 0$ and $x + 3y + 1 = 0$ and perpendicular to the line $7x + 4y - 10 = 0$.

6. The base of a triangle is the line segment from (3,0) to (2,-3). If the area of the triangle is 7, find the general equation of the locus of the third vertex. (Two solutions).

7. Find the distance between the given lines:

a. $3x - y = 5$ and $3x - y = 25$

b. $4x - 6y = 9$ and $2x - 3y = 6$

Solution:

1) Find the line through $(\frac{1}{2}, 3)$ and $(0,0)$. The equation of the line: $y = ax + b$.

$(0,0)$ is in the line, so $0 = a*0 + b$. then $b = 0$. And $(\frac{1}{2}, 3)$ is in the line, so $3 = \frac{a}{2}$.

Then $a = 6$. We have a line $y = 6x$. The line parallel to this is $y = 6x + c$. It passing through $(1,0)$. So $0 = 6 * 1 + c$. Then $c = -6$.

Answer: $y = 6x - 6$.

2) The line parallel to $3x - 7y - 11 = 0$ is $3x - 7y + d = 0$. It passing through $(2, -4)$. So $3 * 2 - 7(-4) + d = 0$; $6 + 28 + d = 0$. Then $d = -34$.

Answer: $3x - 7y - 34 = 0$.

3) Like in $1^{\text{st}}$ task find the line through $(4, 1)$ and $(-2, 7)$. The equation of the line: $y = ax + b$.

$$\left\{ \begin{array}{l} 1 = a * 4 + b \\ 7 = a * (-2) + b \end{array} \right.$$

From the $1^{\text{st}}$ equation $b = 1 - 4a$. Replace to the $2^{\text{nd}}$ equation.

$$7 = -2a + (1 - 4a);$$ $$7 = -6a + 1;$$ $$6 = -6a;$$

$a = -1$. And $b = 1 - 4(-1) = 1 + 4 = 5$. So the line $y = -x + 5$. It is equivalent to $+y - 5 = 0$. The line perpendicular to this is $x - y + d = 0$.

It passing through $\left(-\frac{1}{3}, \frac{1}{2}\right)$. So $-\frac{1}{3} - \frac{1}{2} + d = 0$, $d = \frac{5}{6}$.

Answer: $x - y + \frac{5}{6} = 0$.

4) The line with slope $-\frac{5}{2}$ is $y = -\frac{5}{2}x + c$. It is equivalent to $\frac{5}{2}x + y - c = 0$. The line perpendicular to this is $x - \frac{5}{2}y + d = 0$. It passing through $\left(-\frac{1}{2}, 2\right)$.

So $-\frac{1}{2} - \frac{5}{2} * 2 + d = 0, d = 5 + \frac{1}{2} = \frac{11}{2}$.

Answer: $x - \frac{5}{2}y + \frac{11}{2} = 0$.

5) The point of intersection of lines $5x - 2y - 12 = 0$ and $x + 3y + 1 = 0$ is

$$\left\{ \begin{array}{l} 5x - 2y - 12 = 0 \\ x + 3y + 1 = 0 \end{array} \right.$$

Solve this system. From the $2^{\text{nd}}$ equation $x = -3y - 1$. Replace to the $1^{\text{st}}$ equation.

$$\begin{array}{l} 5(-3y - 1) - 2y - 12 = 0; \\ -15y - 5 - 2y - 12 = 0; \\ -17y - 17 = 0; \\ \end{array}$$

$y = -1$. So $x = -3(-1) - 1 = 2$. So the point of intersection is $(2, -1)$.

The line perpendicular to the line $7x + 4y - 10 = 0$ is $7x + 4y + d = 0$. And through the point $(2, -1)$ is $14 - 4 + d = 0$, $d = -10$.

Answer: $7x + 4y - 10 = 0$.

6) The base of a triangle is the line segment from $(3,0)$ to $(2,-3)$. If the area of the triangle is 7, find the general equation of the locus of the third vertex. (Two solutions).

The line through $(3,0)$ and $(2,-3)$ is $y = ax + b$.

$$\left\{ \begin{array}{l} 0 = a * 3 + b \\ -3 = a * 2 + b \end{array} \right.$$

From the $1^{\text{st}}$ equation $b = -3a$. Replace to the $2^{\text{nd}}$ equation.

$$\begin{array}{l} -3 = 2a - 3a; \\ -3 = -a; \\ \end{array}$$

$a = 3$. And $b = -3 * 3 = -9$.

So the line is $y = 3x - 9$. It is equivalent to $3x - y - 9 = 0$. (1)

The $3^{\text{rd}}$ vertex is on the line perpendicular to line (1) and through the middle of base.

The middle of the base is the point with coordinates: $= \frac{3 + 2}{2} = \frac{5}{2}$; $y = \frac{0 - 3}{2} = \frac{-3}{2}$.

The line perpendicular to line (1) is $-x + 3y + d = 0$. And through the point $\left(\frac{5}{2}, -\frac{3}{2}\right)$ is

$$\begin{array}{l} -\frac{5}{2} + 3\left(\frac{-3}{2}\right) + d = 0; \\ -\frac{5}{2} - \frac{9}{2} + d = 0; \\ \end{array}$$

$d = 7$.

The $3^{\text{rd}}$ vertex is on the line $-x + 3y + 7 = 0$. Denote the vertex of triangle as $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$, where $(x_1, y_1) = (3,0)$ and $(x_2, y_2) = (2, -3)$. The formula of area of the triangle is

$Area = \frac{1}{2}(x_1y_2 + x_2y_3 + x_3y_1 - x_2y_1 - x_3y_2 - x_1y_3)$. From the task $Area = 7$.

So $7 = \frac{1}{2}(-9 + 2y_3 + 0 - 0 - x_3(-3) - 3y_3)$.

$$\begin{array}{l} 14 = -9 - y_3 + 3x_3; \\ 3x_3 - y_3 = 23. \\ \end{array}$$

And $-x_3 + 3y_3 + 7 = 0$.

We have a system

$$\left\{ \begin{array}{c} 3 x _ {3} - y _ {3} = 2 3 \\ - x _ {3} + 3 y _ {3} + 7 = 0 \end{array} \right.$$

From the $2^{\text{nd}}$ equation $x_{3} = 3y_{3} + 7$ . Replace to the $1^{\text{st}}$ equation.

$$3 (3 y _ {3} + 7) - y _ {3} = 2 3;$$ $$9 y _ {3} + 2 1 - y _ {3} = 2 3;$$ $$8 y _ {3} = 2.$$ $$y _ {3} = \frac {1}{4}. x _ {3} = \frac {3}{4} + 7 = \frac {3 1}{4}.$$

So the third vertex is $C\left(\frac{31}{4}, \frac{1}{4}\right)$ . Another solution is the point $D(x, y)$ in the line is on the line $-x + 3y + 7 = 0$ , and symmetric about the middle of base $M\left(\frac{5}{2}, -\frac{3}{2}\right)$ .

From the formula of the middle of line segment: $\frac{5}{2} = \frac{\frac{31}{4} + x}{2}$ ; $-\frac{3}{2} = \frac{\frac{1}{4} + y}{2}$ .

$$5 = \frac {3 1}{4} + x; - 3 = \frac {1}{4} + y.$$ $$x = - \frac {1 1}{4}; y = - \frac {1 3}{4}.$$

Answer: $\left(\frac{31}{4},\frac{1}{4}\right)$ and $\left(-\frac{11}{4}, - \frac{13}{4}\right)$ .

7)

When the lines are given by

$$a x + b y + c _ {1} = 0$$ $$a x + b y + c _ {2} = 0,$$

the distance between them can be expressed as

$$d = \frac {\left| c _ {2} - c _ {1} \right|}{\sqrt {a ^ {2} + b ^ {2}}}.$$

a) Lines $3x - y = 5$ and $3x - y = 25$ . The distance is equal $\frac{|5 - 25|}{\sqrt{3^2 + (-1)^2}} = \frac{20}{\sqrt{10}}$ .

Answer: $\frac{20}{\sqrt{10}}$

b) Lines $4x - 6y = 9$ and $2x - 3y = 6$ . The second equation is equivalent to

$$4 x - 6 y = 1 2. \text {The distance is equal} \frac {| 9 - 1 2 |}{\sqrt {4 ^ {2} + (- 6) ^ {2}}} = \frac {3}{\sqrt {1 6 + 3 6}} = \frac {3}{\sqrt {5 2}}$$

Answer: $\frac{3}{\sqrt{52}}$

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