Co-ordinate Geometry: Solved Examples

Example 1. In which quadrant does the point (-3,4) lie?

Sol: Option 2
The point is negative in the x axis and positive for the y axis , thus the point must lie in the 2nd quadrant.

Example 2. Find the co-ordinates of a point on x-axis, which is at a distance of 5 units from the point (6, -3).

Sol: Option 1
Let the co-ordinates of the point on the x-axis be (x, 0).
As distance =&redic;(x₂ - x₁)² + (y₂ - y₁)²,
so 5 = &redic;(x-6)² + (0-(-3))²
⇒ 25 = x2 - 12x + 36 + 9.
⇒ x = 2 or x = 10
So the required points are (2, 0) and (10, 0).

Example 3. The quadrants where abcissa and ordinate have different signs are?

Sol: Option 2
The signs are different for 2nd and 4th quadrants.

Example 4. Find the co-ordinates of the point of intersection of the medians of triangle MNO; given M = (-2, 3), N = (6, 7), O = (4, 1).

Sol: Option 3
Point of intersection of medians is called as centroid, at which each median is divided in the ratio 2 : 1. Firstly find the midpoint of NO and take it as P

Take the point C, where medians meet and it will divide the median MP in the ratio 2: 1.

Example 5. What is the slope of the line passing through the points J (-2, 3) and (2, 7)?

Sol: Option 2

Example 6. Find the equation of the line whose slope is 2 and y intercept is - 3.

Sol: Option 3
Given m = 2 and c = - 3. Substituting values in y = mx + c, we get y = 2x – 3.

Example 7. Find the equation of the line passing through (2, -1) and parallel to the line 2x – y = 4.

Sol: Option 3
The given line is 2x – y = 4 ⇒ y = 2x – 4 (Converting into the form of y = mx + c)
Its slope = 2. The slope of the parallel line should also be 2.
Hence for the required line
m = 2 and (x1 , y1) = (2, -1).
Equation = (y-y¹) / (x/x¹) = (y²-y¹) / (x²-x¹)
⇒ (y-y¹) / (x/x¹) = m
⇒ y-y¹ = m (x-x¹) ⇒ y - (-1) = 2(x-2)
⇒ y = 2x – 5.

Example 8. Find the coordinate of the point which will divide the line joining the point (2, 4) and (7,9) internally in the ratio 1:2?

Sol: Option 4
The internal division will use the formula (mx₂ + nx₁ )/ (m + n)
y = (my₂ + ny₁ )/(m + n).
So, the point becomes (11/3,17/3).

Example 9. Find the coordinates of the point which will divide the line joining the points (3, 5) and (11, 8) externally in the ratio 5: 2.

Sol: Option 4
The external division case will use the formula
x = (mx₂ – nx₁ )/(m – n)
y = (my₂ – ny₁ )/ (m – n)
where m:n is 5:2 in our case.
Putting the values you will get points (49/3, 10).

Example 10. Find the equation of straight line passing through (2, 3) and perpendicular to the line 3x + 2y + 4 = 0

Sol: Option 2
The given line is 3x + 2y + 4 = 0 or y = (-3x/2) - 2
Any line perpendicular to it will have slope = 2/3
Thus equation of line through (2, 3) and slope 2/3 is
(y – 3) = 2/3 (x – 2)
⇒ 3y – 9 = 2x – 4
⇒ 3y – 2x – 5 = 0.

Example 11. Find the area of the triangle formed by the vertices (4, 5), (10, 12) and (-3, 2)

Sol: Option 4
The area of the triangle having its vertices as P(x1, y1), Q(x2, y2) and R(x3, y3) is given by