Class 8-12 Mathematics Practice Questions, Solutions | ICSE ISC CBSE Math Mathematics Made Easy for School Students

$\displaystyle \textbf{Question 1. }\text{Write the distance of the point }P(2,3,5)\text{ from the }xy\text{-plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Distance from }xy\text{-plane is }|z|$
$\displaystyle \therefore \text{Distance}=|5|=5$
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$\displaystyle \textbf{Question 2. }\text{Write the distance of the point }P(3,4,5)\text{ from }z\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Distance from }z\text{-axis}=\sqrt{x^2+y^2}$
$\displaystyle =\sqrt{3^2+4^2}$
$\displaystyle =5$
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$\displaystyle \textbf{Question 3. }\text{If the distance between the points }P(a,2,1)\text{ and }Q(1,-1,1)\text{ is } \\ 5\text{ units, find the value of }a.$
$\displaystyle \text{Answer:}$
$\displaystyle PQ=\sqrt{(a-1)^2+(2+1)^2+(1-1)^2}$
$\displaystyle 5=\sqrt{(a-1)^2+9}$
$\displaystyle 25=(a-1)^2+9$
$\displaystyle (a-1)^2=16$
$\displaystyle a-1=\pm4$
$\displaystyle \therefore a=5\text{ or }-3$
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$\displaystyle \textbf{Question 4. }\text{The coordinates of the mid-points of sides }AB,BC\text{ and }CA\text{ of }\triangle ABC\text{ are}$
$\displaystyle D(1,2,-3),\ E(3,0,1)\text{ and }F(-1,1,-4)\text{ respectively. Write the coordinates of its centroid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Centroid of }\triangle ABC\text{ is same as centroid of the medial triangle }DEF$
$\displaystyle G=\left(\frac{1+3-1}{3},\frac{2+0+1}{3},\frac{-3+1-4}{3}\right)$
$\displaystyle =\left(1,1,-2\right)$
$\displaystyle \therefore \text{Centroid}=(1,1,-2)$
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$\displaystyle \textbf{Question 5. }\text{Write the coordinates of the foot of the perpendicular from the point } \\ (1,2,3)\text{ on }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Any point on }y\text{-axis is of the form }(0,y,0)$
$\displaystyle \therefore \text{Foot of perpendicular}=(0,2,0)$
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$\displaystyle \textbf{Question 6. }\text{Write the length of the perpendicular drawn from the point } \\ P(3,5,12) \text{ on }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Distance from }x\text{-axis}=\sqrt{y^2+z^2}$
$\displaystyle =\sqrt{5^2+12^2}$
$\displaystyle =\sqrt{169}=13$
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$\displaystyle \textbf{Question 7. }\text{Write the coordinates of third vertex of a triangle having centroid}(3,-5,7)$
$\displaystyle \text{at the origin and two vertices at and }(3,0,1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the third vertex be }(x,y,z)$
$\displaystyle \text{Centroid}=\left(\frac{3+3+x}{3},\frac{-5+0+y}{3},\frac{7+1+z}{3}\right)=(0,0,0)$
$\displaystyle \frac{6+x}{3}=0,\quad \frac{-5+y}{3}=0,\quad \frac{8+z}{3}=0$
$\displaystyle x=-6,\quad y=5,\quad z=-8$
$\displaystyle \therefore \text{Third vertex}=(-6,5,-8)$
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$\displaystyle \textbf{Question 8. }\text{What is the locus of a point }(x,y,z)\text{ for which }y=0,\ z=0?$
$\displaystyle \text{Answer:}$
$\displaystyle y=0,\ z=0$
$\displaystyle \text{Hence only }x\text{ varies.}$
$\displaystyle \therefore \text{The locus is the }x\text{-axis.}$
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$\displaystyle \textbf{Question 9. }\text{Find the ratio in which the line segment joining the points }(2,4,5)$
$\displaystyle \text{ and } (3,-5,4) \text{is divided by the }yz\text{-plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the }yz\text{-plane divide the line joining }A(2,4,5)\text{ and }B(3,-5,4) \\ \text{ in the ratio }m:n$
$\displaystyle \text{The }yz\text{-plane is }x=0$
$\displaystyle \text{Using section formula, }x\text{-coordinate of dividing point is}$
$\displaystyle \frac{3m+2n}{m+n}=0$
$\displaystyle 3m+2n=0$
$\displaystyle 3m=-2n$
$\displaystyle \therefore m:n=-2:3$
$\displaystyle \text{Hence the }yz\text{-plane divides the line externally in the ratio }2:3$
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$\displaystyle \textbf{Question 10. }\text{Find the point on }y\text{-axis which is at a distance of }\sqrt{10} \text{ units from} \\ \text{the point }(1,2,3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the point on }y\text{-axis be }(0,y,0)$
$\displaystyle \text{Distance from }(1,2,3)\text{ is }\sqrt{10}$
$\displaystyle \sqrt{(0-1)^2+(y-2)^2+(0-3)^2}=\sqrt{10}$
$\displaystyle 1+(y-2)^2+9=10$
$\displaystyle (y-2)^2=0$
$\displaystyle y=2$
$\displaystyle \therefore \text{Required point is }(0,2,0)$
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$\displaystyle \textbf{Question 11. }\text{Find the point on }x\text{-axis which is equidistant from the points } \\ A(3,2,2)\text{ and }B(5,5,4).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point on }x\text{-axis be }(x,0,0)$
$\displaystyle \text{Since it is equidistant from }A\text{ and }B,$
$\displaystyle (x-3)^2+(0-2)^2+(0-2)^2=(x-5)^2+(0-5)^2+(0-4)^2$
$\displaystyle (x-3)^2+8=(x-5)^2+41$
$\displaystyle x^2-6x+9+8=x^2-10x+25+41$
$\displaystyle -6x+17=-10x+66$
$\displaystyle 4x=49$
$\displaystyle x=\frac{49}{4}$
$\displaystyle \therefore \text{Required point is }\left(\frac{49}{4},0,0\right)$
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$\displaystyle \textbf{Question 12. }\text{Find the coordinates of a point equidistant from the origin and points } \\ A(a,0,0),B(0,b,0)\text{ and }C(0,0,c).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point be }P(x,y,z)$
$\displaystyle PO^2=PA^2=PB^2=PC^2$
$\displaystyle x^2+y^2+z^2=(x-a)^2+y^2+z^2$
$\displaystyle \Rightarrow x=\frac{a}{2}$
$\displaystyle x^2+y^2+z^2=x^2+(y-b)^2+z^2$
$\displaystyle \Rightarrow y=\frac{b}{2}$
$\displaystyle x^2+y^2+z^2=x^2+y^2+(z-c)^2$
$\displaystyle \Rightarrow z=\frac{c}{2}$
$\displaystyle \therefore P=\left(\frac{a}{2},\frac{b}{2},\frac{c}{2}\right)$
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$\displaystyle \textbf{Question 13. }\text{Write the coordinates of the point }P\text{ which is five-sixth of the way from } \\ A(-2,0,6)\text{ to }B(10,-6,-12).$
$\displaystyle \text{Answer:}$
$\displaystyle P=A+\frac{5}{6}(B-A)$
$\displaystyle B-A=(10+2,-6-0,-12-6)=(12,-6,-18)$
$\displaystyle P=(-2,0,6)+\frac{5}{6}(12,-6,-18)$
$\displaystyle =(-2,0,6)+(10,-5,-15)$
$\displaystyle =(8,-5,-9)$
$\displaystyle \therefore P=(8,-5,-9)$
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$\displaystyle \textbf{Question 14. }\text{If a parallelepiped is formed by the planes drawn through the points} \\ (2,3,5)\text{ and }(5,9,7) \text{parallel to the coordinate planes, then write the lengths of edges of} \\ \text{the parallelepiped and length of the diagonal.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Lengths of edges are differences of corresponding coordinates.}$
$\displaystyle |5-2|=3,\quad |9-3|=6,\quad |7-5|=2$
$\displaystyle \therefore \text{Lengths of edges are }3,6,2$
$\displaystyle \text{Length of diagonal}=\sqrt{3^2+6^2+2^2}$
$\displaystyle =\sqrt{9+36+4}$
$\displaystyle =7$
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$\displaystyle \textbf{Question 15. }\text{Determine the point on }yz\text{-plane which is equidistant from points }$
$\displaystyle A(2,0,3),B(0,3,2) \text{ and }C(0,0,1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point on }yz\text{-plane be }P(0,y,z)$
$\displaystyle PA^2=PB^2=PC^2$
$\displaystyle PA^2=(0-2)^2+y^2+(z-3)^2$
$\displaystyle =4+y^2+z^2-6z+9=y^2+z^2-6z+13$
$\displaystyle PB^2=y^2-6y+9+z^2-4z+4$
$\displaystyle =y^2+z^2-6y-4z+13$
$\displaystyle PC^2=y^2+(z-1)^2=y^2+z^2-2z+1$
$\displaystyle PA^2=PC^2$
$\displaystyle y^2+z^2-6z+13=y^2+z^2-2z+1$
$\displaystyle -4z=-12$
$\displaystyle z=3$
$\displaystyle PB^2=PC^2$
$\displaystyle y^2+z^2-6y-4z+13=y^2+z^2-2z+1$
$\displaystyle -6y-2z+12=0$
$\displaystyle 3y+z=6$
$\displaystyle 3y+3=6$
$\displaystyle y=1$
$\displaystyle \therefore P=(0,1,3)$
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$\displaystyle \textbf{Question 16. }\text{If the origin is the centroid of a triangle }ABC\text{ having vertices } \\ A(a,1,3),B(-2,b,-5)$
$\displaystyle \text{and }C(4,7,c),\text{ find the values of }a,b,c.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since origin is the centroid,}$
$\displaystyle \left(\frac{a-2+4}{3},\frac{1+b+7}{3},\frac{3-5+c}{3}\right)=(0,0,0)$
$\displaystyle \frac{a+2}{3}=0,\quad \frac{b+8}{3}=0,\quad \frac{c-2}{3}=0$
$\displaystyle a=-2,\quad b=-8,\quad c=2$
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