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Learn/ 9th Mathematics/ Unit 7 /Numericals

Unit 7: Coordinate Geometry — Exercise 7 1

9th Class Mathematics · Unit 7: Coordinate Geometry

7.1.2.(i) Find the distance between the points: A(6, 7), B(0, -2) (ii) Find the distance between the points: C(-5, -2), D(3, 2) (iii) Find the distance between the points: L(0, 3), M(-2, -4) (iv) Find the distance between the points: P(-8, -7), Q(0, 0)
Given
(i) Points A(6,7), B(0,-2)
Formula
(i) Distance formula d=|overline{AB}|=sqrt{(x2-x1)2+(y2-y1)2}
Working
(i) Substituting =sqrt{(0-6)2+(-2-7)2}
(i) =sqrt{(-6)2+(-9)2}=sqrt{36+81}=sqrt{117}
Result
(i) Distance d=3sqrt{13}
Given
(ii) Points C(-5,-2), D(3,2)
Formula
(ii) Distance formula d=|overline{CD}|=sqrt{(x2-x1)2+(y2-y1)2}
Working
(ii) Substituting =sqrt{(3-(-5))2+(2-(-2))2}
(ii) =sqrt{(8)2+(4)2}=sqrt{64+16}=sqrt{80}
Result
(ii) Distance d=4sqrt{5}
Given
(iii) Points L(0,3), M(-2,-4)
Formula
(iii) Distance formula d=|overline{LM}|=sqrt{(x2-x1)2+(y2-y1)2}
Working
(iii) Substituting =sqrt{(-2-0)2+(3-(-4))2}
(iii) =sqrt{(-2)2+(7)2}=sqrt{4+49}
Result
(iii) Distance d=sqrt{53}
Given
(iv) Points P(-8,-7), Q(0,0)
Formula
(iv) Distance formula d=|overline{PQ}|=sqrt{(x2-x1)2+(y2-y1)2}
Working
(iv) Substituting =sqrt{(0-(-8))2+(0-(-7))2}
(iv) =sqrt{(8)2+(7)2}=sqrt{64+49}
Result
(iv) Distance d=sqrt{113}
7.1.3.(a) Find (i) the distance between the two given points (ii) midpoint of the line segment joining the two points: A(3, 1), B(-2, -4) (b) Find (i) the distance between the two given points (ii) midpoint of the line segment joining the two points: A(-8, 3), B(2, -1) (c) Find (i) the distance between the two given points (ii) midpoint of the line segment joining the two points: A(-√5, -1/3), B(-3√5, 5)
Given
(a) Points A(3,1), B(-2,-4)
Formula
(a) Distance |AB|=sqrt{(-2-3)2+(-4-1)2}
(a) =sqrt{25+25}=sqrt{50}=5sqrt{2}
(a) Midpoint Midpoint of AB=left(3-22,1-42right)
Result
(a) Results |AB|=5sqrt{2}, Midpoint=left(tfrac12,-tfrac32right)
Given
(b) Points A(-8,3), B(2,-1)
Formula
(b) Distance |overline{AB}|=sqrt{(2-(-8))2+(-1-3)2}
(b) =sqrt{100+16}=sqrt{116}=2sqrt{29}
(b) Midpoint Midpoint=left(-8+22,3-12right)=(-3,1)
Result
(b) Results |AB|=2sqrt{29}, Midpoint=(-3,1)
Given
(c) Points A(-sqrt5,-tfrac13), B(-3sqrt5,5)
Formula
(c) Distance |AB|=sqrt{(-3sqrt5+sqrt5)2+(5+tfrac13)2}
(c) =sqrt{(2sqrt5)2+left(163right)2}=sqrt{20+2569=sqrt{4369=tfrac{2sqrt{109}{3}
(c) Midpoint Midpoint=left(-sqrt5-3sqrt52,-tfrac13+52right)=(-2sqrt5,tfrac73)
Result
(c) Results |AB|=tfrac{2sqrt{109}{3}, Midpoint=(-2sqrt5,tfrac73)
7.1.4.(i) Which of the following points are at a distance of 15 units from the origin? (√176, 7) (ii) Which of the following points are at a distance of 15 units from the origin? (10, -10) (iii) Which of the following points are at a distance of 15 units from the origin? (1, 15)
Formula
(i) Distance from origin sqrt{(sqrt{176}-0)2+(7-0)2}
(i) =sqrt{176+49}=sqrt{225}=15
Result
(i) the point (sqrt{176},7) is 15 units from origin
Formula
(ii) Distance from origin sqrt{(10-0)2+(-10-0)2}
(ii) =sqrt{100+100}=sqrt{200}=10sqrt2neq15
Result
(ii) the point (10,-10) is not 15 units from origin
Formula
(iii) Distance from origin sqrt{(1-0)2+(15-0)2}
(iii) =sqrt{1+225}=sqrt{226}
Result
(iii) the point (1,15) is not 15 units from origin
7.1.5.(i) Show that the points A(0, 2), B(√3, 1) and C(0, -2) are vertices of a right triangle. (ii) Show that the points A(3, 1), B(-2, -3) and C(2, 2) are vertices of an isosceles triangle. (iii) Show that the points A(5, 2), B(-2, 3), C(-3, -4) and D(4, -5) are vertices of a parallelogram.
Formula
(i) Distance formula d=sqrt{(x2-x1)2+(y2-y1)2}
(i) AB AB=sqrt{(sqrt3-0)2+(1-2)2}=sqrt4=2
(i) BC BC=sqrt{(0-sqrt3)2+(-2-1)2}=sqrt{12}
(i) AC AC=sqrt{(0-0)2+(-2-2)2}=4
(i) Pythagoras (AC)2=(AB)2+(BC)2
(i) (4)2=(2)2+(sqrt{12})2 ⇒ 16=4+12=16
Result
(i) the given points are vertices of a right triangle
Given
(ii) Points A(3,1), B(-2,-3), C(2,2)
(ii) AB |AB|=sqrt{(-2-3)2+(-3-1)2}=sqrt{25+16}=sqrt{41}
(ii) BC |BC|=sqrt{(2-(-2))2+(2-(-3))2}=sqrt{16+25}=sqrt{41}
(ii) CA |CA|=sqrt{(3-2)2+(1-2)2}=sqrt{1+1}=sqrt2
Result
(ii) |AB|=|BC| ⇒ A,B,C are vertices of an isosceles triangle
Given
(iii) Points A(5,2), B(-2,3), C(-3,-4), D(4,-5)
(iii) AB |AB|=sqrt{(-7)2+(1)2}=sqrt{50}=5sqrt2
(iii) BC |BC|=sqrt{(-1)2+(-7)2}=sqrt{50}=5sqrt2
(iii) CD |CD|=sqrt{(7)2+(-1)2}=sqrt{50}=5sqrt2
(iii) DA |DA|=sqrt{(1)2+(7)2}=sqrt{50}=5sqrt2
Result
(iii) |AB|=|CD|, |BC|=|DA| ⇒ A,B,C,D are vertices of a parallelogram
7.1.6.Find h such that the points A(√3, -1), B(0, 2) and C(h, -2) are vertices of a right triangle with right angle at the vertex A.
Formula
Pythagoras (right angle at A) |AB|2+|CA|2=|BC|2
Working
Substituting left[(0-sqrt3)2+(2+1)2right]+left[(sqrt3-h)2+(-1+2)2right]=(h-0)2+(-2-2)2
[3+9]+[3-2sqrt3h+h2+1]=h2+16 ⇒ 12+4-2sqrt3h+h2=h2+16
-2sqrt3h=h2+16-12-4-h2=0 ⇒ -2sqrt3h=0
Result
h=0
7.1.7.Find h such that A(-1, h), B(3, 2) and C(7, 3) are collinear.
Formula
Collinearity determinant begin{vmatrix}x1&y1&1x2&y2&1x3&y3&1end{vmatrix}=0
Working
Substituting begin{vmatrix}-1&h&13&2&17&3&1end{vmatrix}=0
-1(2-3)-h(3-7)+1(9-14)=0 ⇒ 1+4h-5=0
4h-4=0 ⇒ 4h=4
Result
h=1
7.1.8.The points A(-5, -2) and B(5, -4) are ends of a diameter of a circle. Find the centre and radius of the circle.
Formula
Centre = midpoint of AB C=left(-5+52,-2-42right)
=left(02,-62right)=(0,-3)
Radius = |AC| r=sqrt{(0+5)2+(-3+2)2}
=sqrt{25+1}
Result
Centre=(0,-3), r=sqrt{26}
7.1.9.Find h such that the points A(h, 1), B(2, 7) and C(-6, -7) are vertices of a right triangle with right angle at the vertex A.
Formula
Pythagoras |AB|2+|AC|2=|BC|2
Working
Distances |AB|=sqrt{40-4h+h2}; |BC|=sqrt{260}; |AC|=sqrt{h2+12h+100}
h2+4h-60=0 ⇒ (h+10)(h-6)=0
Result
h=-10 or h=6
7.1.10.A quadrilateral has the points A(9, 3), B(-7, 7), C(-3, -7) and D(5, -5) as its vertices. Find the midpoints of its sides. Show that the figure formed by joining the midpoints consecutively is a parallelogram.
Given
Vertices A(9,3), B(-7,7), C(-3,-7), D(5,-5)
Midpoints E=(1,5), F=(-5,0), G=(1,-6), H=(7,-1)
Side lengths of EFGH |EF|=sqrt{61}, |FG|=6sqrt2, |GH|=sqrt{61}, |HE|=6sqrt2
Result
|EF|=|GH| and |FG|=|HE| ⇒ EFGH is a parallelogram