This is the Multiple Choice Questions Part 3 of the Series in Geometry topic in Engineering Mathematics. In Preparation for the ECE Board Exam make sure to expose yourself and familiarize in each and every questions compiled here taken from various sources including but not limited to past Board Examination Questions in Engineering Mathematics, Mathematics Books, Journals and other Mathematics References.
MCQ Topic Outline included in Mathematics Board Exam Syllabi
- MCQ in Lines and Planes | MCQ in lane figures | MCQ in Application of Cavalier’s, Pappus and Prismodial Theorems | MCQ in Coordinate in Space | MCQ in Quadratic Surfaces | MCQ in Mensuration | MCQ in Plane Geometry | MCQ in Solid Geometry | MCQ in Spherical Geometry | MCQs in Analytical Geometry
Continue Practice Exam Test Questions Part 3 of the Series
⇐ MCQ in Geometry Part 2 | Mathematics Board Exam
Choose the letter of the best answer in each questions.
101. Find the distance between the lines 3x + y – 12 = 0 and 3x + y – 4 = 0.
A. 16/sqrt(10)
B. 12/sqrt(10)
C. 4/sqrt(10)
D. 8/sqrt(10)
Answer: Option D
Solution:
102. What is the length of the line with a slope of 4/3 from a point (6, 4) to the y-axis?
A. 10
B. 25
C. 50
D. 75
Answer: Option A
Solution:
103. Find the slope of the line defined by y – x = 5.
A. 1
B. 1/4
C. -1/2
D. 5 + x
Answer: Option A
Solution:
104. What is the slope of the line 3x + 2y + 1 = 0?
A. 3/2
B. 2/3
C. -3/2
D. -2/3
Answer: Option C
Solution:
105. In a Cartesian coordinates, the vertices of a triangle are defined by the following points: (-2, 0), (4, 0) and (3, 3). What is the area?
A. 8 sq. units
B. 9 sq. units
C. 10 sq. units
D. 11 sq. units
Answer: Option B
Solution:
106. Given three vertices of a triangle whose coordinates are A (1, 1), B (3, -3) and (5, -3). Find the area of the triangle.
A. 3
B. 4
C. 5
D. 6
Answer: Option B
Solution:
107. In a Cartesian coordinates, the vertices of a square are: (1, 1), (0, 8) (4, 5) and (-3, 4). What is the area?
A. 20 sq. units
B. 30 sq. units
C. 25 sq. units
D. 35 sq. units
Answer: Option C
Solution:
108. A line passes thru (1, -3) and (-4, 2). Write the equation of the line in slope-initial form
A. y – 4 = x
B. y = -x – 2
C. y = x – 4
D. y – 2 = x
Answer: Option B
Solution:
109. What is the x-intercept of the line passing through (1, 4) and (4, 1)?
A. 4.5
B. 5
C. 4
D. 6
Answer: Option B
Solution:
110. Find the equation of a straight line with a slope of 3 and a y-intercept of 1.
A. 3x + y – 1 = 0
B. 3x – y + 1 = 0
C. x + 3y + 1 = 0
D. x – 3y – 1 = 0
Answer: Option B
Solution:
111. If the points (-2, 3), (x, y) and (-3, 5) lie on a straight line, then the equation of the line is _________.
A. x – 2y – 1 = 0
B. 2x + y – 1 = 0
C. x + 2y – 1 = 0
D. 2x + y + 1 = 0
Answer: Option D
Solution:
112. The equation of a line that intercepts the x-axis at x = 4 and the y-axis at y = -6 is,
A. 3x + 2y = 12
B. 2x – 3y = 12
C. 3x – 2y = 12
D. 2x – 3y = 12
Answer: Option C
Solution:
113. A line with an inclination of 45° passes through (-5/2, 9/2). What is the x-coordinate of a point on the line if its corresponding y-coordinate is 6?
A. 6
B. 7
C. 8
D. 9
Answer: Option C
Solution:
114. Find the equation of the line passing through the origin and with a slope of 6?
A. y – 6x = 0
B. y = -6
C. x + y = -6
D. 6x + y = 0
Answer: Option A
Solution:
115. Find the equation of the line if the x-intercept and y-intercept are -2 and 4, respectively.
A. y – 2x – 4 = 0
B. y + 2x – 4 = 0
C. y – 2x + 4 = 0
D. y + 2x + 4 = 0
Answer: Option B
Solution:
116. Determine B such that 3x + 2y – 7 = 0 is perpendicular to 2x – By + 2 = 0.
A. 5
B. 4
C. 3
D. 2
Answer: Option C
Solution:
117. The line 2x – 3y + 2 = 0 is perpendicular to another line L1 of unknown equation. Find the slope of L1.
A. 3/2
B. -3/2
C. 2/3
D. -2/3
Answer: Option B
Solution:
118. A line through (-5, 2) and (1, -4) is perpendicular to the line through (x, -7) and (8, 7). Find x.
A. -4
B. -5
C. -6
D. -19/3
Answer: Option C
Solution:
119. What is the equation of the line that passes through (4, 0) and is parallel to the line x – y – 2 = 0?
A. x – y + 4 = 0
B. x + y + 4 = 0
C. x – y – 4 = 0
D. x – y = 0
Answer: Option C
Solution:
120. Find the equation of the line through point (3, 1) and is perpendicular to the line x + 5y + 5 = 0.
A. 5x – 2y = 14
B. 5x – y = 14
C. 2x – 5y = 14
D. 2x + 5y = 14
Answer: Option B
Solution:
121. Find the equation of the perpendicular bisector of the line joining (5, 0) and (-7, 3)
A. 8x + 2y + 11 = 0
B. 8x – 2y + 11 = 0
C. 8x – y + 11 = 0
D. 8x + y + 11 = 0
Answer: Option B
Solution:
122. Which of the following lines is parallel to the line 3x – 2y + 6 = 0?
A. 3x + 2y – 12 = 0
B. 4x – 9y = 6
C. 12x + 18y = 15
D. 15x – 10y – 9 = 0
Answer: Option D
Solution:
123. The equation of the line through (-3, 5) parallel to 7x + 2y -4 = 0 is
A. 7x + 2y + 31 = 0
B. 7x – 2y + 30 = 0
C. 7x + 2y – 4 = 0
D. 2x + 7y + 30 = 0
Answer: Option A
Solution:
124. What is the equation of the line joining the points (3, -2) and (-7, 6)?
A. 2x + 3y = 0
B. 4x – 5y = 22
C. 4x + 5y = 2
D. 5x + 4y = 7
Answer: Option C
Solution:
125. What is the equation of the line passing through (-2, 6) with the x-intercept half the y-intercept?
A. x – y = 6
B. 2x + 2y + 2 = 0
C. 3x – y + 2 = 0
D. 2x + y – 2 = 0
Answer: Option D
Solution:
126. Find the slope of a line having a parametric equation of x = 2 + t and y = 5 – 3t.
A. 2
B. 3
C. -2
D. -3
Answer: Option D
Solution:
127. Find the slope of the line having a parametric equation y = 4t + 6 and x = t + 1.
A. 1
B. 2
C. 3
D. 4
Answer: Option D
Solution:
128. Two vertices of a triangle are (2, 4) and (-2, 3) and the area is 2 square units, the locus of the third vertex is
A. 4x – y = 14
B. 4x + 4y = 14
C. x + 4y = 12
D. x – 4y = -14
Answer: Option D
Solution:
129. Find the area of the triangle which the line 2x – 3y + 6 = 0 forms with the coordinate axis.
A. 3
B. 4
C. 5
D. 2
Answer: Option A
Solution:
130. A line passes through point (2, 2). Find the length equation of the line if the length of the line segment intercepted by the coordinates axed is the square root of 5.
A. 2x + y – 2 = 0
B. 2x – y – 2 = 0
C. 2x – y + 2 = 0
D. 2x + y + 2 = 0
Answer: Option B
Solution:
131. What is the radius of the circle x2 + y2 – 6y = 0?
A. 2
B. 3
C. 4
D. 5
Answer: Option B
Solution:
132. What are the coordinates of the center of the curve x2 + y2 – 2x – 4y – 31 = 0?
A. (-1, -1)
B. (-2, -2)
C. (1, 2)
D. (2, 1)
Answer: Option C
Solution:
133. A circle whose equation is x2 + y2 + 4x + 6y – 23 = 0 has its center at:
A. (2, 3)
B. (3, 2)
C. -3, 2)
D. (-2, -3)
Answer: Option D
Solution:
134. What is the radius of a circle with the ff. Equation: x2 – 6x + y2 – 4y – 12 = 0
A. 3.46
B. 7
C. 5
D. 6
Answer: Option C
Solution:
135. The diameter of a circle described by 9x2 + 9y2 = 16 is
A. 4/3
B. 16/9
C. 8/3
D. 4
Answer: Option C
Solution:
136. How far from the y-axis is the center of the curve 2x2 + 2y2 + 10x – 6y – 55 = 0?
A. -2.5
B. -3.0
C. -2.75
D. -3.25
Answer: Option A
Solution:
137. What is the distance between the centers of the circles x2 + y2 + 2x + 4y – 3 = 0 and x2 + y2 -8x -6y + 7 = 0?
A. 7.07
B. 7.77
C. 8.07
D. 7.87
Answer: Option A
Solution:
138. The shortest distance from A(3, 8) to the circle x2 + y2 + 4x – 6y = 12 is equal to
A. 2.1
B. 2.3
C. 2.5
D. 2.7
Answer: Option B
Solution:
139. The equation x2 + y2 – 4x + 2y – 20 = 0 describes:
A. A circle of radius 5 centered at the origin.
B. An ellipse centered at (2, -1)
C. A sphere centered at the origin.
D. A circle of radius 5 centered at (2, -1)
Answer: Option D
Solution:
140. The center of a circle is at (1, 1) and one point on its circumference is (-1, -3.) Find the other end of the diameter through (-1, -3).
A. (2, 4)
B. (3, 5)
C. (3, 6)
D. (1, 3)
Answer: Option B
Solution:
141. Find the area (in square units) of the circle whose equation is x2 + y2 = 6x – 8y.
A. 20π
B. 22π
C. 25π
D. 27π
Answer: Option C
Solution:
142. Determine the equation of the circle whose radius is 5, center on the line x = 2 and tangent to the line 3x – 4y + 11 = 0.
A. (x – 2)2 + (y – 2)2 = 5
B. (x – 2)2 + (y + 2)2 = 25
C. (x – 2)2 + (y + 2)2 = 5
D. (x – 2)2 + (y – 2)2 = 25
Answer: Option B
Solution:
143. Find the equation of the circle with the center at (-4, 5) and tangent to the line 2x + 7y – 10 = 0.
A. x2 + y2 + 8x – 10y – 12 = 0
B. x2 + y2 + 8x – 10y + 12 = 0
C. x2 + y2 + 8x + 10y – 12 = 0
D. x2 + y2 + 8x + 10y + 12 = 0
Answer: Option C
Solution:
144. Find the value of k for which the equation x2 + y2 + 4x – 2y – k = 0 represents a point circle.
A. 5
B. 6
C. -6
D. -5
Answer: Option A
Solution:
145. 3x2 + 2x – 5y + 7 = 0. Determine the curve.
A. Parabola
B. Ellipse
C. Circle
D. Hyperbola
Answer: Option A
Solution:
146. The focus of the parabola y2 = 4x is at:
A. (4, 0)
B. (0, 4)
C. (3, 0)
D. (0, 3)
Answer: Option A
Solution:
147. Where is the vertex of the parabola x2 = 4(y – 2)?
A. (2, 0)
B. (0, 2)
C. (3, 0)
D. (0, 3)
Answer: Option B
Solution:
148. Find the equation of the directrix of the parabola y2 = 16x.
A. x = 2
B. x = -2
C. x = 4
D. x = -4
Answer: Option D
Solution:
149. Given the equation of a parabola 3x + 2y2 – 4y + 7 = 0. Locate its vertex.
A. (5/3, 1)
B. (5/3, -1)
C. (-5/3, -1)
D. (-5/3, 1)
Answer: Option D
Solution:
150. In the equation y = -x2 + x +1, where is the curve facing?
A. upward
B. facing left
C. facing right
D. downward
Answer: Option D
Solution:
Online Question and Answer in Geometry Series
Following is the list of multiple choice questions in this brand new series:
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