Point and Straight Line Quiz-2
Dear Readers,
Dear Readers,
As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. The equation of the straight line perpendicular to 5x-2y=7 and passing through the point of intersection of the lines 2x+3y=1 and 3x+4y=6, is
Solution
a) Let the equation of line which is perpendicular to 5x-2y=7, is 2x+5y=λ …(i) The point of intersection of given lines is (14,-9) Since, the Eq. (i) is passing through the point (14,-9) ∴ 2(14)+5(-9)=λ ⇒ λ=-17 ∴ Eq. (i) becomes 2x+5y+17=0
a) Let the equation of line which is perpendicular to 5x-2y=7, is 2x+5y=λ …(i) The point of intersection of given lines is (14,-9) Since, the Eq. (i) is passing through the point (14,-9) ∴ 2(14)+5(-9)=λ ⇒ λ=-17 ∴ Eq. (i) becomes 2x+5y+17=0
Q2.Let the base of a triangle lie along the line x=a and be of length 2a. The area of this triangle is a2 if the vertex
lies on the lines
lies on the lines
Solution
(b) Let (h,k) be the coordinates of the vertex. Then, the height of the triangle is the length of the perpendicular from (h,k) on x=a i.e. |h-a| Since the area of the triangle is a2 ∴1/2 (2 a)|h-a|=a2 ⇒|h-a|=a ⇒h-a=±a ⇒h=0,h=2 a Hence, the vertex lies on x=0 or, x=2a
(b) Let (h,k) be the coordinates of the vertex. Then, the height of the triangle is the length of the perpendicular from (h,k) on x=a i.e. |h-a| Since the area of the triangle is a2 ∴1/2 (2 a)|h-a|=a2 ⇒|h-a|=a ⇒h-a=±a ⇒h=0,h=2 a Hence, the vertex lies on x=0 or, x=2a
Q3. The image of the point (-1,3) by the line x-y=0 is
Solution
a) Let the image of the point (-1,3) in the line y=x is (3,-1)
a) Let the image of the point (-1,3) in the line y=x is (3,-1)
Q4. The angle between the pair of straight lines formed by joining the points of intersection of x2+y2=4 and y=3x+c to the origin is a right angle. Then, c2 is equal to
Solution
(a) Since, the angle is right angle. ∴Homogenising,x2+y2=4((y-3x)/c)2 ⇒ c2 (x2+y2)=4(y2+9x2-6xy) Since, lines are at right angle. ∴ Coefficient of x2+coefficient of y2=0 ⇒ c2-36+c2-4=0 ⇒ c2=20
(a) Since, the angle is right angle. ∴Homogenising,x2+y2=4((y-3x)/c)2 ⇒ c2 (x2+y2)=4(y2+9x2-6xy) Since, lines are at right angle. ∴ Coefficient of x2+coefficient of y2=0 ⇒ c2-36+c2-4=0 ⇒ c2=20
Q5. One vertex of the equilateral triangle with centroid at the origin and one side as x+y-2=0 is
Solution
(c) Let ABC be the equilateral triangle with centroid O(0,0) and sides BC as x+y-2=0. ∴OD=|(0+0-2)/√(12+12)|=√2⇒OA=2√2 Since AD is perpendicular to BC. Therefore, Slope of AD=1 ⇒AD makes 45° with X-axis Clearly, A lies on OA at a distance of 2√2 units from O. So, its coordinates are given by (x-0)/cos〖Ï€/4〗 =(y-0)/sin〖Ï€/4〗 =±2√2 ⇒x=±2,y=±2 But, O and A lie on the same side of x+y-2=0 Hence, the coordinates of A are (-2,-2)
(c) Let ABC be the equilateral triangle with centroid O(0,0) and sides BC as x+y-2=0. ∴OD=|(0+0-2)/√(12+12)|=√2⇒OA=2√2 Since AD is perpendicular to BC. Therefore, Slope of AD=1 ⇒AD makes 45° with X-axis Clearly, A lies on OA at a distance of 2√2 units from O. So, its coordinates are given by (x-0)/cos〖Ï€/4〗 =(y-0)/sin〖Ï€/4〗 =±2√2 ⇒x=±2,y=±2 But, O and A lie on the same side of x+y-2=0 Hence, the coordinates of A are (-2,-2)
Q6. If the lines 4x+3y-1=0,x-y+5=0 and kx+5y-3=0 are concurrent, then k is equal to
Solution
c) Since, the given three lines are concurrent, then ⇒ 4(3-25)-3(-3-5k)-1(5+k)=0 ⇒ -88+9+15k-5-k=0 ⇒ 14k=84 ⇒ k=6
c) Since, the given three lines are concurrent, then ⇒ 4(3-25)-3(-3-5k)-1(5+k)=0 ⇒ -88+9+15k-5-k=0 ⇒ 14k=84 ⇒ k=6
Q7.(-5,0) and B(3,0) are two of the vertices of a triangle ABC. Its area is 20 square cms. The
vertex C lies on the line x-y=2. The coordinates of C are
vertex C lies on the line x-y=2. The coordinates of C are
Solution
(d) Let the coordinates of the third vertex C be (h,k). Then, Area of ABC=20 sq.units Since, (h,k) lies on x-y=2 Therefore, h-k=2 …(ii) Solving (i) and (ii), we find that the coordinates of the third vertex are (-3,-5) or,(7,5)
(d) Let the coordinates of the third vertex C be (h,k). Then, Area of ABC=20 sq.units Since, (h,k) lies on x-y=2 Therefore, h-k=2 …(ii) Solving (i) and (ii), we find that the coordinates of the third vertex are (-3,-5) or,(7,5)
Q8.The circumcentre of the triangle formed by the lines xy+2 x+2y+4=0 and x+y+2=0, is
Solution
(c) We have, xy+2 x+2 y+4=0 ⇒(x+2)(y+2)=0⇒x+2=0,y+2=0 Solving the equations of the sides of the triangle we obtain the coordinates of the vertices as A(-2,0),B(0,-2)C(-2,-2). Clearly, ∆ ABC is a right angled triangle with right angle at C. Therefore, the centre of the circumcircle is the mid-point of AB whose coordinates are (-1,-1)
(c) We have, xy+2 x+2 y+4=0 ⇒(x+2)(y+2)=0⇒x+2=0,y+2=0 Solving the equations of the sides of the triangle we obtain the coordinates of the vertices as A(-2,0),B(0,-2)C(-2,-2). Clearly, ∆ ABC is a right angled triangle with right angle at C. Therefore, the centre of the circumcircle is the mid-point of AB whose coordinates are (-1,-1)
Q9. The image of the origin with reference to the line 4x+3y-25=0 is
Solution
b) Let the image of reflection of the origin with reference to the line 4x+3y-25=0 is (h,k) ∴(h-0)/4=(k-0)/3=(-2(0+0-25))/(16+9)=2 ⇒ h/4=2 ⇒ h=8 and k/3=2 ⇒ k=6 ∴ Required point is (8, 6)
b) Let the image of reflection of the origin with reference to the line 4x+3y-25=0 is (h,k) ∴(h-0)/4=(k-0)/3=(-2(0+0-25))/(16+9)=2 ⇒ h/4=2 ⇒ h=8 and k/3=2 ⇒ k=6 ∴ Required point is (8, 6)
Q10. Consider the family of lines (x+y-1)+λ(2x+3y-5)=0 and
(3x+2y-4)+μ(x+2y-6)=0, equation of the straight line that belongs to both the families is
Solution
(b) The family of lines (x+y-1)+λ(2x+3y-5)=0 passes through a point such that x+y-1=0 2x+3y-5=0 ie,(-2,3) and family of lines (3x+2y-4)+μ(x+2y-6)=0 Passes through a point such that 3x+2y-4=0 and x+2y-6=0 ie,(-1,7/2) ∴ Equation of the straight line that belongs to both the families passes through (-2,3) and (-1,7/2) is y-3=(7/2-3)/(-1+2) (x+2) ⇒y-3=(x+2)/2⇒x-2y+8=0
(b) The family of lines (x+y-1)+λ(2x+3y-5)=0 passes through a point such that x+y-1=0 2x+3y-5=0 ie,(-2,3) and family of lines (3x+2y-4)+μ(x+2y-6)=0 Passes through a point such that 3x+2y-4=0 and x+2y-6=0 ie,(-1,7/2) ∴ Equation of the straight line that belongs to both the families passes through (-2,3) and (-1,7/2) is y-3=(7/2-3)/(-1+2) (x+2) ⇒y-3=(x+2)/2⇒x-2y+8=0
