Coordinate Geometry Questions and Answers

Answer: B

To find the slope, we rearrange the equation into the slope-intercept form (y = mx + c).

5x + 2y = 7

2y = -5x + 7

y = \(-\frac{5}{2}x + \frac{7}{2}\)

The slope (m) is the coefficient of x, which is -5/2.

Answer: A

The points of trisection divide the line segment into three equal parts. The point closer to A(3, -4) will divide the segment AB in the ratio 1:2.

Let A=(3,-4) and B=(1,2). The ratio is \(m_1:m_2 = 1:2\).

x = \(\frac{1(1) + 2(3)}{1+2} = \frac{1+6}{3} = \frac{7}{3}\)

y = \(\frac{1(2) + 2(-4)}{1+2} = \frac{2-8}{3} = \frac{-6}{3} = -2\)

The coordinates are (7/3, -2).

Answer: D

Correct answer: Isosceles

Compute the three side lengths with the distance formula.

\[(0,6)\text{ to }(-5,3): \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}\]

\[(0,6)\text{ to }(5,3): \sqrt{5^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}\]

\[(-5,3)\text{ to }(5,3): \sqrt{10^2 + 0^2} = 10\]

Two sides are equal and the third differs, so the triangle is isosceles but not equilateral. It is not right-angled either: \(34 + 34 = 68\), which is not \(10^2 = 100\).

Answer: D

We find the lengths of the three sides using the distance formula.

Side 1 (between (0,6) and (-5,3)): \(\sqrt{(-5-0)^2 + (3-6)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}\).

Side 2 (between (-5,3) and (3,2)): \(\sqrt{(3-(-5))^2 + (2-3)^2} = \sqrt{8^2 + (-1)^2} = \sqrt{64+1} = \sqrt{65}\).

Side 3 (between (3,2) and (0,6)): \(\sqrt{(0-3)^2 + (6-2)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\).

Since all three sides have different lengths (\(\sqrt{34}, \sqrt{65}, 5\)), the triangle is scalene.

Answer: A

First, find the slope (m₁) of the line through (2, 5) and (-3, 6).

m₁ = \(\frac{6-5}{-3-2} = \frac{1}{-5} = -1/5\).

The slope (m₂) of the required perpendicular line is the negative reciprocal of m₁. m₂ = 5.

Now use the point-slope form with point (-3, 5) and slope m₂=5.

y - 5 = 5(x - (-3))

y - 5 = 5(x + 3)

y - 5 = 5x + 15

5x - y + 20 = 0.

Answer: A

For the points to be collinear, the slope between the first two points must equal the slope between the second and third points.

Slope between (1, 7) and (2, 4) = \(\frac{4-7}{2-1} = \frac{-3}{1} = -3\).

Slope between (2, 4) and (k, 1) = \(\frac{1-4}{k-2} = \frac{-3}{k-2}\).

Set the slopes equal: -3 = \(\frac{-3}{k-2}\).

This implies k-2 = 1, so k = 3.

Answer: A

Using the section formula:

x = \(\frac{2(4) + 1(1)}{2+1} = \frac{8+1}{3} = 3\)

y = \(\frac{2(6) + 1(3)}{2+1} = \frac{12+3}{3} = 5\)

The point is P(3, 5). Since both coordinates are positive, the point lies in the First Quadrant.

Answer: A

The perpendicular bisector passes through the midpoint of AB and has a slope that is the negative reciprocal of the slope of AB.

1. Find the midpoint of AB: \((\frac{1+3}{2}, \frac{2+8}{2}) = (2, 5)\).

2. Find the slope of AB: \(m_{AB} = \frac{8-2}{3-1} = \frac{6}{2} = 3\).

3. Find the slope of the perpendicular bisector: \(m_{perp} = -1/3\).

4. Use the point-slope form with the midpoint (2, 5) and slope -1/3:

y - 5 = -1/3 (x - 2)

3(y - 5) = -1(x - 2)

3y - 15 = -x + 2

x + 3y = 17.

Answer: B

If the right angle is at B, then the sides AB and BC are perpendicular. The product of their slopes must be -1.

Slope of AB = \(\frac{3-7}{p-4} = \frac{-4}{p-4}\).

Slope of BC = \(\frac{3-3}{7-p} = \frac{0}{7-p} = 0\).

A slope of 0 means BC is a horizontal line. For AB to be perpendicular to a horizontal line, it must be a vertical line. A vertical line has an undefined slope, which happens when its denominator is 0.

So, p - 4 = 0, which means p = 4.

Answer: B

We can find the area by dividing the quadrilateral into two triangles, e.g., ABC and ADC, and summing their areas. Let A=(-4,-2), B=(-3,-5), C=(3,-2), D=(2,3).

Area(\(\triangle ABC\)) = \(\frac{1}{2} |-4(-5-(-2)) -3(-2-(-2)) + 3(-2-(-5))| = \frac{1}{2} |-4(-3) -3(0) + 3(3)| = \frac{1}{2}|12+9| = 10.5\).

Area(\(\triangle ADC\)) = \(\frac{1}{2} |-4(-2-3) + 3(3-(-2)) + 2(-2-(-2))| = \frac{1}{2} |-4(-5) + 3(5) + 2(0)| = \frac{1}{2}|20+15| = 17.5\).

Total Area = 10.5 + 17.5 = 28 sq units.