Question

Find the equation of a line in point - slope form that goes through the points (4, -4) and (3, -6).

Answer 1

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Answer 2

$\large\underline{\bold{Given \:Question - }}$

• Find the equation of a line in point - slope form that goes through the points (4, -4) and (3, -6).

$\large\underline{\bold{ANSWER-}}$

$\large\underline{\bold{Given- }}$

• A line that goes through the points (4, - 4) and (3, - 6).

$\large\underline{\bold{To \: Find - }}$

• The equation of line using point - slope form.

$\large\underline{\bold{Concept \: Used-}}$

1. The slope of line joining the points (a, b) and (c, d) is

$\sf \: slope \: of \: line \: (m) = \dfrac{d - b}{c - a}$

2. Equation of line having slope 'm' and which passes through the point (a, b) is given by

$\sf \: y - b \: = \: m \: (x - a)$

$\large\underline{\bold{Solution-}}$

Given that,

• A line that goes through the points (4, - 4) and (3, - 6).

So,

• Slope of line (m) is evaluated as

$\rm :\longmapsto\:m \: = \: \dfrac{ - 6 + 4}{3 - 4}$

$\rm :\longmapsto\:m \: = \: \dfrac{ - 2}{ - 1}$

$\rm :\implies\:m \: = \: 2$

Now,

• The equation of line having slope 2 and which passes through the point (4, - 4) is given

$\rm :\longmapsto\:y - ( - 4) = 2(x - 4)$

$\rm :\longmapsto\:y + 4 = 2x - 8$

$\rm :\longmapsto\:2x - y - 12 = 0$

Additional Information

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

• Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

• Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line.

2. Point-slope form equation of line.

• Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a)

3. Slope-intercept form equation of line.

• Consider a line whose slope is m which cuts the Y-axis at a distance ‘a’ from the origin. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a.

4. Intercept Form of Line.

• Consider a line L having x– intercept a and y– intercept b, then the line passes through  X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1.

5. Normal form of Line.

• Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, the equation of line is given by x cosβ + y sinβ = p.