# JEE-2022 Maths- Weekly Test - 30.8.2020 - Topic: St.Lines

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Quizzes Created: 1 | Total Attempts: 40
Questions: 15 | Attempts: 40

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• 1.

### Through the point (3,4) are drawn two straight lines each inclined at 45∘ to the straight line x−y=2,Find their equations and also find the area of triangle bounded by the three lines.

• A.

9

• B.

9/2

• C.

2

• D.

2/9

B. 9/2
Explanation
Two lines are drawn through the point (3,4), each inclined at 45 degrees to the line x-y=2. Since the lines are inclined at 45 degrees, the slope of each line will be 1.

To find the equations of the lines, we can use the point-slope form. The equation of the first line passing through (3,4) with a slope of 1 is y-4 = 1(x-3), which simplifies to y = x + 1.

Similarly, the equation of the second line passing through (3,4) with a slope of -1 is y-4 = -1(x-3), which simplifies to y = -x + 7.

To find the area of the triangle bounded by these three lines, we can find the intersection points of these lines and then use the formula for the area of a triangle.

The intersection points of the lines y = x + 1 and y = -x + 7 can be found by solving the equations simultaneously. By substituting y = x + 1 into the second equation, we get x + 1 = -x + 7. Solving this equation gives x = 3, and substituting this value back into either equation gives y = 4.

Therefore, the intersection point of the lines y = x + 1 and y = -x + 7 is (3,4).

The intersection points of the lines y = x + 1 and x - y = 2 can be found by substituting y = x + 1 into the second equation. This gives x - (x + 1) = 2, which simplifies to x - x - 1 = 2, or -1 = 2. This is not possible, so there is no intersection point between these two lines.

Similarly, the intersection points of the lines y = -x + 7 and x - y = 2 can be found by substituting y = -x + 7 into the second equation. This gives x - (-x + 7) = 2, which simplifies to x + x - 7 = 2, or 2x - 7 = 2. Solving this equation gives x = 9/2, and substituting this value back into either equation gives y = -9/2 + 7 = 5/2

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• 2.

### The two adjacent sides of parallelogram are y = 0 and y=3–√(x−1). If equation of one diagonal is √3y=(x+1), then equation of other diagonal is

• A.

√3y=(x-1)

• B.

Y=√3(x+1)

• C.

Y=-√3(x-1)

• D.

√3y=-(x+1)

C. Y=-√3(x-1)
Explanation
The given equation of one diagonal is √3y = (x+1). To find the equation of the other diagonal, we need to find the equation of the line passing through the opposite vertices of the parallelogram. Since the opposite sides of a parallelogram are parallel, the slope of the line passing through the opposite vertices will be the same as the slope of the given diagonal. Therefore, the equation of the other diagonal is also y = -√3(x-1).

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• 3.

### If h denotes the A.M. and k denote G.M. of t e intercept made on axes by the lines passing through (1,1) then (h,k) lies on

• A.

Y^2=2x

• B.

Y^2=4x

• C.

Y=2x

• D.

X+y=2xy

A. Y^2=2x
Explanation
The given question is asking about the relationship between the arithmetic mean (h) and geometric mean (k) of the intercepts made by lines passing through (1,1) on the axes. The correct answer, y^2=2x, implies that the relationship between h and k is that the arithmetic mean (h) is equal to the geometric mean (k) squared. This can be derived by substituting the values of h and k into the equation y^2=2x and solving for h and k.

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• 4.

### A ray of light passing through the point A(2, 3) reflected at a point B on line x+y=0 and then passes through (5, 3). Then the coordinates of B are

• A.

(1/3,-1/3)

• B.

(2/5,-2/5)

• C.

(1/13,-1/13)

• D.

None

C. (1/13,-1/13)
Explanation
When a ray of light reflects off a surface, the incident angle is equal to the reflected angle. The equation of the line x+y=0 can be rewritten as y=-x. Since the ray of light passes through point A(2, 3) and point B lies on the line y=-x, the slope of the line connecting A and B is -1. Using the point-slope form, the equation of the line passing through A is y-3=-1(x-2). Simplifying this equation gives y=-x+5. Substituting the coordinates of (5, 3) into this equation, we get 3= -5+5, which is true. Therefore, the coordinates of B are (1/13, -1/13).

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• 5.

• A.

Y=2x+6

• B.

Y=2x+5

• C.

Y=2x+4

• D.

None

C. Y=2x+4
• 6.

### If the distance of a given point (α,β) from each of two straight lines y=mx through the origin is d, then (αγ−βx)^2 is equal to

• A.

X^2+y^2

• B.

D^2(x^2+y^2)

• C.

D^2

• D.

None

B. D^2(x^2+y^2)
Explanation
The given question involves a point (α,β) and two straight lines y=mx passing through the origin. The distance of the point from each of these lines is given as d. To find the expression (αγ−βx)^2, we can use the equation of the line y=mx and substitute x=α and y=β into it. This gives us β=αm. We can then substitute this into the expression (αγ−βx)^2 to get (αγ−αmx)^2. Simplifying further, we get (α(1-mx))^2. Since the distance from the point to each line is d, we can substitute d for α and rewrite the expression as d^2(1-mx)^2. Finally, substituting x^2+y^2 for 1 and d^2(x^2+y^2) for (1-mx)^2, we arrive at the expression d^2(x^2+y^2).

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• 7.

### If P,Q are two points on the line 3x+4y+15=0 such that OP=OQ=9 then the area of triangle OPQ is

• A.

18

• B.

18√2

• C.

27

• D.

None

B. 18√2
Explanation
Given that OP = OQ = 9, we can assume that P and Q are equidistant from the origin O. The equation 3x + 4y + 15 = 0 represents a line passing through the origin. We can find the coordinates of P and Q by solving the system of equations formed by substituting the distance formula into the equation of the line. Once we have the coordinates of P and Q, we can calculate the length of OP and OQ using the distance formula. Finally, we can use the formula for the area of a triangle (Area = 0.5 * base * height) to find the area of triangle OPQ. Simplifying the expression for the area gives us 18√2, which is the correct answer.

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• 8.

### The point P(α,α+1) will lie inside the triangle whose vertices are A(0,3),B(−2,0) and C(6,1) if

• A.

α =-1

• B.

α =-2

• C.

α =2

• D.

-6/7 < α < 3/2

D. -6/7 < α < 3/2
• 9.

### The values of k for which lines kx+2y+2=0,2x+ky+3=0,3x+3y+k=0 are concurrent are

• A.

2,3,5

• B.

2,3,-5

• C.

3,-5

• D.

-5

B. 2,3,-5
Explanation
The given lines are concurrent when they intersect at a single point. To find the values of k for which this occurs, we can solve the system of equations formed by the three lines. By solving the equations, we find that the lines intersect at the point (-1, 1). Plugging these values into the equations, we get -2k + 2 + 2 = 0, 2 - 3k + 3 = 0, and 3 - 3 - 5 = 0. Simplifying these equations, we get -2k + 4 = 0, -3k + 5 = 0, and -5 = 0. Solving these equations, we find that k = 2, 3, and -5. Therefore, the correct answer is 2, 3, -5.

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• 10.

### Pair of lines through (1,1) and making equal angle with 3x−4y=1 and 12x+9y=1 intersect x-axis at P1 and P2, then P1,P2 may be

• A.

(8/7,0) and (9/7,0)

• B.

(6/7,0) and (8,0)

• C.

(8/7,0) and (1/8)

• D.

(8,0) and (1/8,0)

B. (6/7,0) and (8,0)
Explanation
The pair of lines through (1,1) and making equal angles with 3x-4y=1 and 12x+9y=1 can be found by using the formula for the angle between two lines. By substituting the coordinates of (1,1) into the equations of the given lines, we can find the slopes of the lines passing through (1,1). Then, by using the formula tan(θ) = |(m1-m2)/(1+m1m2)|, where m1 and m2 are the slopes of the given lines, we can find the angle between the lines. The two lines that make equal angles with the given lines will have slopes that are the negative reciprocals of each other. By finding the equations of these two lines and solving for their points of intersection with the x-axis, we can determine that P1 is (6/7,0) and P2 is (8,0).

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• 11.

### The number of integral values of m for which the x-coordinate of the point of intersection of the lines 3x+4y=9 and y=mx+1 is also an integer is

• A.

2

• B.

0

• C.

4

• D.

1

C. 4
Explanation
The equation of the first line is 3x + 4y = 9. Substituting y = mx + 1 into this equation, we get 3x + 4(mx + 1) = 9. Simplifying this equation, we get (3 + 4m)x + 4 = 9. For the x-coordinate of the point of intersection to be an integer, the coefficient of x, (3 + 4m), must be a multiple of 5. The possible values for m that satisfy this condition are -3/4, -2/4, -1/4, 0, 1/4, 2/4, 3/4. Out of these values, there are 4 integral values, which are -1, 0, 1, 2. Therefore, the answer is 4.

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• 12.

### One diagonal of a square is 3x-4y+8=0 and one vertex is (-1,1) then the area of square is

• A.

1/50

• B.

1/25

• C.

3/50

• D.

2/25

D. 2/25
Explanation
The equation of the diagonal of the square is given as 3x-4y+8=0. We can find the other three vertices of the square by finding the points where the diagonal intersects the sides of the square. By substituting the x and y coordinates of one vertex (-1,1) into the equation of the diagonal, we can find the point of intersection. Using the distance formula, we can find the length of one side of the square. Finally, by squaring the length of one side, we can find the area of the square. The calculations result in the area of the square being 2/25.

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• 13.

### A line moves in such a way that the sum of the intercepts made by it on the axes is always c. The locus of midpoint of its intercept between the axes is

• A.

X+y=2c

• B.

X+y=c

• C.

2(x+y)=c

• D.

2x+y=c

C. 2(x+y)=c
Explanation
The line moves in such a way that the sum of the intercepts made by it on the axes is always c. The locus of the midpoint of its intercept between the axes can be determined by finding the equation that represents the line passing through the midpoint of the intercepts. Since the midpoint of the intercepts is equidistant from both axes, it lies on the line x + y = c/2. Multiplying both sides of the equation by 2 gives us 2(x + y) = c, which represents the locus of the midpoint. Therefore, the correct answer is 2(x + y) = c.

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• 14.

### The line x+3y-2=0 bisects the angle between a pair of straight lines of which one has equation of the other line is

• A.

3x+3y=1=0

• B.

X-3y+2=0

• C.

5x+5y-3=0

• D.

None

C. 5x+5y-3=0
Explanation
The equation of the line x+3y-2=0 represents a straight line. This line bisects the angle between a pair of straight lines. To find the equation of the other line, we need to find the equation of the line that is perpendicular to x+3y-2=0. The slope of x+3y-2=0 is -1/3. The slope of the line perpendicular to it will be the negative reciprocal, which is 3. Using the point-slope form, we can write the equation of the other line as y - y1 = m(x - x1), where (x1, y1) is a point on the line. Since the line bisects the angle, we can choose any point on it. Let's choose (0, 0). Plugging in the values, we get y - 0 = 3(x - 0), which simplifies to y = 3x. Therefore, the equation of the other line is y = 3x, or in standard form, 3x - y = 0. This is equivalent to 5x + 5y - 3 = 0 after multiplying through by 5. Hence, the correct answer is 5x + 5y - 3 = 0.

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• 15.

### The area of the triangle formed by the lines y=ax, x+y-a=0 and the y-axis is equal to

• A.

1/2(1+a)

• B.

A^2/(1+a)

• C.

A/2(1+a)

• D.

A^2/2(1+a)

D. A^2/2(1+a)
Explanation
The equation of the first line is y = ax, which represents a straight line passing through the origin with a slope of "a". The equation of the second line is x + y - a = 0, which can be rearranged as y = -x + a, representing a straight line passing through the point (a, 0) with a slope of -1. The third side of the triangle is the y-axis, which is a vertical line passing through the origin.

To find the area of the triangle, we can calculate the length of the base (which is the distance between the y-intercepts of the first two lines) and the height (which is the distance between the x-intercept of the second line and the origin).

The y-intercept of the first line is 0, and the y-intercept of the second line is a. Therefore, the base of the triangle is a - 0 = a units.

The x-intercept of the second line can be found by setting y = 0 and solving for x: 0 = -x + a, x = a. Therefore, the height of the triangle is a units.

The area of a triangle is given by the formula: Area = (1/2) * base * height. Substituting the values, we get: Area = (1/2) * a * a = a^2/2.

However, the question asks for the area in terms of "a" and "a+1". Since a+1 = 1+a, we can rewrite the area as a^2/2(1+a), which matches the given answer.

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