Coordinate Geometry Questions and Answers

Answer: A

Let's find the slopes of the lines connecting the points.

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

Slope between (-2,1) and (5,2) = \(\frac{2-1}{5-(-2)} = \frac{1}{7}\).

Slope between (5,2) and (2,-2) = \(\frac{-2-2}{2-5} = \frac{-4}{-3} = \frac{4}{3}\).

The product of the slopes of the first and third lines is \((-\frac{3}{4}) \times (\frac{4}{3}) = -1\). Since the product is -1, these two lines are perpendicular, forming a right angle. Therefore, the triangle is a right-angled triangle.

Answer: A

Use the point-slope form.

\[ y - 6 = -\frac{4}{3}(x + 3) \]

\[ 3y - 18 = -4x - 12 \quad\Longrightarrow\quad 4x + 3y = 6 \]

Checking: at \( (-3, 6) \), \( -12 + 18 = 6 \).

Answer: B

The line 4x - y = 1 has slope 4, so a perpendicular line has slope -1/4. Through (2, -5) its equation is

The option 4x - y + 18 = 0 is parallel to the given line (same slope 4), x - 4y - 18 = 0 has slope 1/4, which is neither, and 4x + 3y - 9 = 0 has slope -4/3. Only x + 4y + 18 = 0 also passes through (2, -5): 2 - 20 + 18 = 0.

Answer: C

The value of (p - q) is 3.

The slope through (4, p) and (0, q) is the change in y over the change in x:

\[ \begin{aligned} \frac{p - q}{4 - 0} &= \frac{3}{4} \\ p - q &= 3 \end{aligned} \]

-3 comes from taking the points in the opposite order for y but not for x. 0 would mean a horizontal line, and 4 confuses the run with the rise.

Answer: D

The line is \(r(\cos\theta - \sin\theta) = 1\).

Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\) into \(y = x - 1\):

\[ \begin{aligned} r\sin\theta &= r\cos\theta - 1 \\ r\cos\theta - r\sin\theta &= 1 \\ r(\cos\theta - \sin\theta) &= 1 \end{aligned} \]

The option with a plus sign is the line \(x + y = 1\), which has slope \(-1\), not \(+1\). The options \(r = \cos\theta\) and \(r = \sin\theta\) are circles through the origin (\(x^{2} + y^{2} = x\) and \(x^{2} + y^{2} = y\)), not straight lines.

Answer: C

9,900 triangles can be constructed.

The x-coordinate can take 10 integer values (\(-4\) to 5) and the y-coordinate 11 values (6 to 16).

• P can be any of the \(10 \times 11 = 110\) lattice points.
• PR is horizontal, so R has the same y as P and a different x: 9 choices.
• The right angle is at P, so PQ is vertical: Q has the same x as P and a different y: 10 choices.

\[ 110 \times 9 \times 10 = 9900 \]

110 counts only the choices of P, and 1,100 forgets one of the factors; 10,000 uses 10 instead of 9 for R, allowing R to coincide with P.

Answer: C

The area of S is 32.

S₁ is the rectangle \(-1 \le x \le 3\), \(-5 \le y \le 1\), with area 4 × 6 = 24.

S₂ is bounded by \(y \ge 1\), \(x \le 3\) and \(y \le x + 2\): a triangle with corners (-1, 1), (3, 1) and (3, 5), with area \(\tfrac{1}{2} \times 4 \times 4 = 8\).

The two regions meet only along the segment y = 1, which has no area, so the union has area 24 + 8 = 32. 26 and 28 undercount the triangle, and 34 counts something twice.

Answer: B

The maximum height is 10 meters.

Differentiate the profile and set the slope to zero:

\[ \begin{aligned} \frac{dy}{dx} &= 2-0.2x = 0 \Rightarrow x = 10 \\ y_{max} &= 2(10)-0.1(10)^{2} = 10\ \text{m} \end{aligned} \]

The second derivative, -0.2, is negative, confirming a maximum. The arch meets the ground at x = 0 and x = 20 m, and the top is midway.

12 and 14 m exceed the largest value the parabola can reach. 8 m is reached at x = 5.53 m and again at x = 14.47 m, on the way up and down, not at the peak.

Answer: A

The enclosed area is 0.5.

The line through (0, 0) and (1, 3) is \(y = 3x\); the line through (0, 0) and (1, 2) is \(y = 2x\). Between \(x = 0\) and \(x = 1\) the first lies above the second, so the area between them is

\[ \begin{aligned} \int_{0}^{1} (3x - 2x)\,dx &= \left[\frac{x^{2}}{2}\right]_{0}^{1} \\ &= 0.5 \end{aligned} \]

Geometrically it is a triangle with vertices (0, 0), (1, 3) and (1, 2): base 1 (the vertical side at \(x = 1\)) and height 1, area \(\frac{1}{2}\). 1.5 and 1.0 are the areas under the individual lines, not between them, and 2.0 matches neither the area between them nor either line.

Answer: B

One intersection point is (0.2, 0.4).

Write the two circles and subtract one equation from the other to get the line through both intersection points:

\[ \begin{aligned} (x - 0.5)^{2} + y^{2} &= 0.25 \Rightarrow x^{2} - x + y^{2} = 0 \\ (x - 1)^{2} + (y - 1)^{2} &= 1 \\ \Rightarrow x^{2} - 2x + y^{2} - 2y + 1 &= 0 \end{aligned} \]

Subtracting gives \(x + 2y - 1 = 0\), so \(x = 1 - 2y\). Substituting in the first circle: \(5y^{2} - 2y = 0\), so \(y = 0\) or \(y = 0.4\). The points are (1, 0) and (0.2, 0.4).

(0, 0) lies on the first circle but is \(\sqrt{2}\) from (1, 1); (0.5, 0.5) and (1, 2) lie on only one circle or neither.