**Answer
1** :

The y-coordinate of every point on the x-axis is 0.

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is x = 0.

**Answer
2** :

We know that the equation of the line passing through point , whose slope is m, is .

Thus, the equation of the line passing through point (–4, 3), whose slope is , is

**Answer
3** :

We know that the equation of the line passing through point , whose slope is m, is .

Thus, the equation of the line passing through point (0, 0), whose slope is m,is

(y – 0) = m(x – 0)

i.e., y = mx

**Answer
4** :

The slope of the line that inclines with the x-axis at an angle of 75° is

m = tan 75°

We know that the equation of the line passing through point , whose slope is m, is .

Thus, if a line passes though and inclines with the x-axis at an angle of 75°, then the equation of the line is given as

**Answer
5** :

It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as

y = m(x – d)

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.

The slope of the line is given as m = –2

Thus, the required equation of the given line is

y = –2 [x – (–3)]

y = –2x – 6

i.e., 2x + y + 6 = 0

**Answer
6** :

It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as

y = mx + c

Here, c = 2 and m = tan 30° .

Thus, the required equation of the given line is

**Answer
7** :

It is known that the equation of the line passing through points (x_{1}, y_{1}) and (x_{2}, y_{2}) is .

Therefore, the equation of the line passing through the points (–1, 1) and

(2, –4) is

**Answer
8** :

If p is the length of the normal from the origin to a line and ω is the angle made by the normal with the positive direction of the x-axis, then the equation of the line is given by xcos ω + y sin ω = p.

Here, p = 5 units and ω = 30°

Thus, the required equation of the given line is

x cos 30° + y sin 30° = 5

**Answer
9** :

It is given that the vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5).

Let RL be the median through vertex R.

Accordingly, L is the mid-point of PQ.

By mid-point formula, the coordinates of point L are given by

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is .

Therefore, the equation of RL can be determined by substituting (x1, y1) = (4, 5) and (x2, y2) = (0, 2).

Hence,

Thus, the required equation of the median through vertex R is .

**Answer
10** :

The slope of the line joining the points (2, 5) and (–3, 6) is

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)

Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

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