# Can anyone please explain how to do this?

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(i) The line is parallel to the y axis. Its equation is x=c (where c is a constant). (7,0) is a point on it. Put this value in the equation to obtain c=7. Therefore, the equation of the line is `x=7` .

(ii) The line is parallel to the x axis. Its equation is y=c (where c is a constant). (0,5) is a point on it. Put this value in the equation to obtain c=5. Therefore, the equation of the line is `y=5` .

(iii) The line passes through the points (0,0) and (2,4). Equation of a line passing through two points `(x_1,y_1)` and `(x_2, y_2)` is:

`y=(y_2-y_1)/(x_2-x_1)x+c` --- (1)

Put the co-ordinates of the two points to obtain `y=(4-0)/(2-0)x+c`

i.e. `y=2x+c`

The line passes through (0, 0). Put the co-ordinates in this equation to get,

`0=2*0+c`

`rArr c=0`

Therefore, the equation of the line takes the form `y=2x`

(iv) The line passes through the points (0,2) and (2,0).

Put the co-ordinates of the two points in eq. (1), to obtain `y=(0-2)/(2-0)x+c`

i.e. `y=-x+c`

The line passes through (0, 2). Put the co-ordinates in this equation to get,

`2=-0+c`

`rArr c=2`

Therefore, the equation of the line takes the form `y=-x+2`

(v) The line passes through the points (-4,-2) and (0,-3).

Put the co-ordinates of the two points in eq. (1), to obtain `y=(-3+2)/(0+4)x+c`

i.e. `y=-1/4x+c`

The line passes through (0, -3). Put the co-ordinates in this equation to get,

`-3=-1/4*0+c`

`rArr c=-3`

Therefore, the equation of the line takes the form `y=-1/4x-3`

`rArr 4y=-x-12`

`rArr x+4y+12=0`

(vi) The line passes through the points (-2,-2) and (0,0).

Put the co-ordinates of the two points in eq. (1), to obtain `y=(0+2)/(0+2)x+c`

i.e. `y=x+c`

The line passes through (0, 0). Put the co-ordinates in this equation to get,

`0=0+c`

`rArr c=0`

Therefore, the equation of the line takes the form `y=x` .

Proceeding on similar lines, equations for all the straight lines shown in the graph can be obtained.

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