Question

Consider the line shown on the graph. Enter the

equation of the line in the form y = mx + b where m is
the slope and b is the y-intercept.
y=

Answer 1

Solution -

From the graph, we concluded that line passes through the points (0, 3) and (1, 4) respectively.

So, slope of the line (m) passing through the points (a, b) and (c, d) is

m = d - b/ c - a

So, on substituting the values, we get

m = 4 - 3 / 1 - 0 = 1/1 = 1

Also, line makes an intercept of 3 units in positive direction of y - axis.

So, c = 3

So, Equation of line is given by

y = mx + c

y = 1 × x + 3

y = x + 3

Alternative Method :-

We know, Equation of line passing through the points (a,b) and (c, d) is given by

y - b = d - b / c - a (x - a)

Here,

a = 0

b = 3

c = 1

d = 4

So, on substituting these values in above result, we get :-

y - 3 = 4 - 3 /1 - 0 ( x - 0 )

y - 3 = 1/1 ( x - 0 )

y - 3 = x

y = x + 3

Different forms of equations of a straight line :-

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is y = b.

Equation of line parallel to y - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m isy - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle B with the positive X-axis is x cosB + y sinB = p.