Coordinate Geometry – Straight Line Quiz - ICSE Class 10 Mathematics
Free practice quiz: Coordinate Geometry – Straight Line (ICSE Class 10 Mathematics). 73 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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Find the equation of the straight line passing through the point (2, 3) and perpendicular to the line $3x - 4y + 5 = 0$.
- 4x + 3y - 17 = 0
- 4x - 3y + 1 = 0
- 3x + 4y - 18 = 0
- 3x - 4y + 2 = 0
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Answer: A. 4x + 3y - 17 = 0
The given line is $3x - 4y + 5 = 0$. The slope of this line is $m_1 = -(\frac{3}{-4}) = \frac{3}{4}$. Since the required line is perpendicular to this line, its slope $m_2$ will be the negative reciprocal of $m_1$. So, $m_2 = -\frac{1}{m_1} = -\frac{1}{3/4} = -\frac{4}{3}$. The equation of the line passing through $(2, 3)$ with slope $-\frac{4}{3}$ is given by $y - 3 = -\frac{4}{3}(x - 2)$. Multiplying by 3, we get $3(y - 3) = -4(x - 2)$, which simplifies to $3y - 9 = -4x + 8$, or $4x + 3y - 17 = 0$.
Determine the equation of the line that bisects the angle between the lines $3x - 4y + 7 = 0$ and $4x - 3y + 16 = 0$ and passes through the origin.
- x + y + 9 = 0
- x - y + 9 = 0
- x + y - 9 = 0
- x - y - 9 = 0
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Answer: B. x - y + 9 = 0
The equations of the angle bisectors of two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are given by $\frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}}$. Here, the lines are $3x - 4y + 7 = 0$ and $4x - 3y + 16 = 0$. $\sqrt{A_1^2 + B_1^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. $\sqrt{A_2^2 + B_2^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. So the equations of the angle bisectors are $\frac{3x - 4y + 7}{5} = \pm \frac{4x - 3y + 16}{5}$. This simplifies to $3x - 4y + 7 = \pm (4x - 3y + 16)$. Case 1: $3x - 4y + 7 = 4x - 3y + 16$. Rearranging gives $x + y + 9 = 0$. Case 2: $3x - 4y + 7 = -(4x - 3y + 16)$. $3x - 4y + 7 = -4x + 3y - 16$. Rearranging gives $7x - 7y + 23 = 0$. The question states that the bisecting line passes through the origin $(0, 0)$. Let's check which bisector passes through the origin. For $x + y + 9 = 0$, substituting $(0, 0)$ gives $0 + 0 + 9 = 9 eq 0$. For $7x - 7y + 23 = 0$, substituting $(0, 0)$ gives $7(0) - 7(0) + 23 = 23 eq 0$. This means that neither of the angle bisectors pass through the origin. This implies an error in the question. Let me assume there's a typo in the constants or the question's requirement of passing through the origin. If the question asked for a line parallel to the angle bisectors, that would be different. Let's re-read the question carefully. 'Determine the equation of the line that bisects the angle between the lines $3x - 4y + 7 = 0$ and $4x - 3y + 16 = 0$ AND passes through the origin.' The wording is precise. There must be a mistake in the original problem statement provided by Sam. Let's try to adjust the constant terms so that one of the bisectors passes through the origin. For the bisector $x+y+C=0$ to pass through the origin, $C$ must be 0. So we want $x+y=0$. This is one of the angle bisectors if the constants are adjusted. Let's re-examine the bisector equations. $3x - 4y + 7 = 4x - 3y + 16 \Rightarrow x + y + 9 = 0$. $3x - 4y + 7 = -4x + 3y - 16 \Rightarrow 7x - 7y + 23 = 0$. If the intention was for the line $x-y+9=0$ to be the answer, its slope is 1. The slopes of the bisectors are -1 and 1. So $x-y+9=0$ is one of the angle bisectors. For this to be the correct option, the second bisector should not pass through the origin, and the first bisector should pass through the origin, or there's a specific reason to choose this one. If the line is $x-y+9=0$, it does not pass through the origin $(0+0+9 eq 0)$. If the line is $x+y+9=0$, it does not pass through the origin $(0+0+9 eq 0)$. The answer provided is $x-y+9=0$. This is one of the angle bisectors. If this is the correct answer, then it must pass through the origin. But it doesn't. The problem statement or the given answer is flawed. Let me check if the other option $x+y+9=0$ is related. No. What if the question meant to ask for one of the angle bisectors and not necessarily passing through the origin? Even then, the question explicitly states it passes through the origin. Let's assume the question is about finding one of the angle bisectors, and the condition 'passes through the origin' is erroneous, and the option $x-y+9=0$ is the intended answer. Let's check if the points given are intended to result in one of these bisectors. $3x - 4y + 7 = 0$ and $4x - 3y + 16 = 0$. Bisectors are $x+y+9=0$ and $7x-7y+23=0$. The option is $x-y+9=0$. This does not match. Let's assume there is a typo in the original equations such that $x-y+9=0$ is indeed an angle bisector AND passes through the origin. If $x-y+9=0$ is an angle bisector and passes through the origin, then $0-0+9=0$, which is false. This question has fundamental flaws. To make the question valid and have $x-y+9=0$ as an answer, and assuming it's an angle bisector and passes through the origin, means $0-0+9$ should be 0. This is impossible. Let me try to make $x-y+9=0$ one of the angle bisectors and it should also pass through the origin. This cannot happen as $0-0+9 eq 0$. Let me assume the question meant to ask for the line that bisects the angle between $3x-4y+7=0$ and $4x-3y+16=0$ AND the line itself is $x-y+9=0$. Then the task is just to verify if $x-y+9=0$ is an angle bisector. It is not. The angle bisectors are $x+y+9=0$ and $7x-7y+23=0$. The option is $x-y+9=0$. This question is completely broken. I will have to skip this question. It's the third unresolvable question.
Find the equation of the line passing through $(2,3)$ and perpendicular to the line $y = 2x - 5$.
- 2x + y - 7 = 0
- x + 2y - 8 = 0
- 2x - y + 1 = 0
- x - 2y + 4 = 0
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Answer: B. x + 2y - 8 = 0
The given line is $y = 2x - 5$. Its slope is $m_1 = 2$. The required line is perpendicular to this line, so its slope $m_2$ is the negative reciprocal of $m_1$. $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$. The line passes through the point $(2, 3)$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 3 = -\frac{1}{2}(x - 2)$. Multiply by 2: $2(y - 3) = -(x - 2) \Rightarrow 2y - 6 = -x + 2$. Rearranging the terms to match the options, we get $x + 2y - 6 - 2 = 0$, which is $x + 2y - 8 = 0$. This matches option 2. The provided answer is option 2, which is correct. No correction needed for this question.
Find the equation of the median from vertex B of triangle ABC, where A=(1, 3), B=(4, 2), and C=(7, 6).
- 5x - 3y - 14 = 0
- 5x + 3y - 26 = 0
- 3x - 5y + 14 = 0
- 3x + 5y - 26 = 0
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Answer: A. 5x - 3y - 14 = 0
A median from vertex B connects B to the midpoint of AC. The midpoint D of AC is $(\frac{1+7}{2}, \frac{3+6}{2}) = (\frac{8}{2}, \frac{9}{2}) = (4, 4.5)$. The line passes through B=(4, 2) and D=(4, 4.5). Since the x-coordinates of B and D are the same, the line is vertical. The equation of a vertical line passing through $(4, 2)$ is $x = 4$. Let me recheck the coordinates. A=(1, 3), B=(4, 2), C=(7, 6). Midpoint of AC is D = $(\frac{1+7}{2}, \frac{3+6}{2}) = (4, \frac{9}{2})$. The median connects B(4,2) to D(4, 9/2). Since both points have the same x-coordinate, the line is $x=4$. None of the options are $x=4$. Let me check the provided answer: $5x - 3y - 14 = 0$. If this is the median, it must pass through B=(4,2). $5(4) - 3(2) - 14 = 20 - 6 - 14 = 14 - 14 = 0$. So it passes through B. Now let's check if the midpoint D=(4, 4.5) lies on this line. $5(4) - 3(4.5) - 14 = 20 - 13.5 - 14 = 6.5 - 14 = -7.5 eq 0$. So the midpoint does not lie on the line. This question is flawed. I will skip it.
Find the equation of the line passing through $(-1, -2)$ and parallel to the line joining $(0, 0)$ and $(3, 4)$.
- 4x - 3y - 5 = 0
- 3x - 4y - 5 = 0
- 4x + 3y + 10 = 0
- 3x + 4y + 11 = 0
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Answer: A. 4x - 3y - 5 = 0
First, find the slope of the line joining the points $(0, 0)$ and $(3, 4)$. The slope is $m_1 = \frac{4-0}{3-0} = \frac{4}{3}$. The required line is parallel to this line, so it will have the same slope, $m_2 = \frac{4}{3}$. The line passes through the point $(-1, -2)$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - (-2) = \frac{4}{3}(x - (-1))$. This becomes $y + 2 = \frac{4}{3}(x + 1)$. Multiply by 3: $3(y + 2) = 4(x + 1) \Rightarrow 3y + 6 = 4x + 4$. Rearranging to match the options, we get $4x - 3y + 4 - 6 = 0$, which is $4x - 3y - 2 = 0$. Let me recheck. Slope of line joining (0,0) and (3,4) is 4/3. Parallel line has slope 4/3. Point is (-1,-2). Equation $y-(-2) = 4/3(x-(-1)) \Rightarrow y+2 = 4/3(x+1) \Rightarrow 3y+6 = 4x+4 \Rightarrow 4x-3y-2=0$. My answer is $4x-3y-2=0$. The provided answer is option 1: $4x - 3y - 5 = 0$. Let's check if $(-1,-2)$ lies on $4x-3y-5=0$. $4(-1) - 3(-2) - 5 = -4 + 6 - 5 = 2 - 5 = -3 eq 0$. So the provided answer is incorrect. My calculated answer $4x-3y-2=0$ is correct. As Tom, I am correcting Sam's provided answer.