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Learn/ 9th Mathematics/ Unit 7 /Numericals

Unit 7: Coordinate Geometry — Exercise 7 2

9th Class Mathematics · Unit 7: Coordinate Geometry

7.2.1.(i) Find the slope and inclination of the line joining the points: (-2, 4); (5, 11) (ii) Find the slope and inclination of the line joining the points: (3, -2); (2, 7) (iii) Find the slope and inclination of the line joining the points: (4, 6); (4, 8)
Formula
(i) Slope m=y2-y1x2-x1
Working
(i) =11-45+2=77=1
Formula
(i) Inclination tanα=m=1
Result
(i) α=tan-1(1)=45^circ
Formula
(ii) Slope m=y2-y1x2-x1
Working
(ii) =7+22-3=9-1=-9
(ii) tanα=-9 ⇒ tan(180-α)=9 ⇒ 180-α=tan-1(9)=83^circ40'
Result
(ii) α=180^circ-83^circ40'=96^circ20'
Formula
(iii) Slope m=y2-y1x2-x1
Working
(iii) =8-64-4=20=infty
Result
(iii) α=tan-1(infty)=90^circ
7.2.2.(i) By means of slopes, show that the following points lie on the same line: A(-1, -3); B(1, 5); C(2, 9) (ii) By means of slopes, show that the following points lie on the same line: P(4, -5); Q(7, 5); R(10, 15) (iii) By means of slopes, show that the following points lie on the same line: L(-4, 6); M(3, 8); N(10, 10) (iv) By means of slopes, show that the following points lie on the same line: X(a, 2b); Y(c, a+b); Z(2c-a, 2a)
(i) Slope of AB 5-(-3)1-(-1)=82=4
(i) Slope of BC 9-52-1=41=4
Result
(i) Slope of AB=Slope of BC ⇒ A,B,C lie on the same line
(ii) Slope of PQ 5-(-5)7-4=103
(ii) Slope of QR 15-510-7=103
(ii) Slope of PQ=Slope of QR ⇒ P,Q,R lie on the same line
(iii) Slope of LM 8-63+4=27
(iii) Slope of MN 10-810-3=27
(iii) Slope of LM=Slope of MN ⇒ L,M,N lie on the same line
(iv) Slope of XY (a+b)-2bc-a=a-bc-a
(iv) Slope of YZ 2a-(a+b)(2c-a)-c=a-bc-a
(iv) Slope of XY=Slope of YZ ⇒ X,Y,Z lie on the same line
7.2.3.Find k so that the line joining A(7, 3); B(k, -6) and the line joining C(-4, 5); D(-6, 4) are: (i) parallel (ii) perpendicular.
Slope of AB m1=-6-3k-7=-9k-7
Slope of CD m2=4-5-6+4=-1-2=frac12
Formula
(i) Parallel condition m1=m2 ⇒ -9k-7=frac12 ⇒ -18=k-7
Result
(i) Result k=-11
Formula
(ii) Perpendicular condition m1m2=-1 ⇒ left(-9k-7right)left(frac12right)=-1 ⇒ -9=-2(k-7)
9=2k-14 ⇒ 2k=23
Result
(ii) Result k=232
7.2.4.Using slopes, show that the triangle with its vertices A(6, 1), B(2, 7) and C(-6, -7) is a right triangle.
Slope of AB m1=7-12-6=6-4=-frac32
Slope of BC m2=-7-7-6-2=-14-8=frac74
Slope of CA m3=1+76+6=812=frac23
m1m3=left(-frac32right)left(frac23right)=-1
Result
Triangle ABC is a right triangle with mangle A=90^circ
7.2.5.(a) Two pairs of points are given. Find whether the two lines determined by these points are (i) parallel (ii) perpendicular (iii) none: (1, -2), (2, 4) and (4, 1), (-8, 2) (b) Two pairs of points are given. Find whether the two lines determined by these points are (i) parallel (ii) perpendicular (iii) none: (-3, 4), (6, 2) and (4, 5), (-2, -7)
(a) Slope of line through (1,-2),(2,4) m1=4+22-1=61=6
(a) Slope of line through (4,1),(-8,2) m2=2-1-8-4=1-12
(a) m1neq m2;quad m1m2=6 · 1-12=-frac12neq-1
Result
(a) lines are neither parallel nor perpendicular
(b) Slope of line through (-3,4),(6,2) m1=2-46+3=-frac29
(b) Slope of line through (4,5),(-2,-7) m2=-7-5-2-4=-12-6=2
(b) m1neq m2;quad m1m2=left(-frac29right)(2)=-frac49neq-1
(b) lines are neither parallel nor perpendicular
7.2.6.(a) Find an equation of the horizontal line through (7, -9). (b) Find an equation of the vertical line through (-5, 3). (c) Find an equation of the line through A(-6, 5) having slope 7. (d) Find an equation of the line through (8, -3) having slope 0. (e) Find an equation of the line through (-8, 5) having slope undefined. (f) Find an equation of the line through (-5, -3) and (9, -1). (g) Find an equation of the line with y-intercept: -7 and slope: -5. (h) Find an equation of the line with x-intercept: -3 and y-intercept: 4. (i) Find an equation of the line with x-intercept: -9 and slope: -4.
7.2.7.Find an equation of the perpendicular bisector of the segment joining the points A(3, 5) and B(9, 8).
Given
Points A(3,5), B(9,8)
Midpoint of AB left(3+92,5+82right)=left(6,132right)
Slope of AB m=8-59-3=frac36=frac12
Slope of perpendicular -1m=-1tfrac12=-2
Formula
Equation of bisector y-132=-2(x-6)
y-132=-2x+12 ⇒ 2x+y-372=0
Result
4x+2y-37=0
7.2.8.Find an equation of the line through (-4, -6) and perpendicular to a line having slope -3/2.
Given
(x1,y1)=(-4,-6), m=-tfrac32
Slope of required line -frac1m=-1-3/2=frac23
Formula
y-(-6)=frac23(x-(-4))
3(y+6)=2(x+4) ⇒ 3y+18=2x+8
Result
2x-3y-10=0
7.2.9.Find an equation of the line through (11, -5) and parallel to a line with slope -24.
Given
(x1,y1)=(11,-5), m=-24
Formula
y-(-5)=-24(x-11)
y+5=-24x+264 ⇒ 24x-264+y+5=0
Result
24x+y-259=0
7.2.10.(a) Convert 2x - 4y + 11 = 0 into slope intercept form, two intercept form and normal form. (b) Convert 4x + 7y - 2 = 0 into slope intercept form, two intercept form and normal form. (c) Convert 15y - 8x + 3 = 0 into slope intercept form, two intercept form and normal form.
(a) (i) Slope-intercept 4y=2x+11 ⇒ y=tfrac12x+114 (m=tfrac12, c=114)
(a) (ii) Two-intercept 2x-4y=-11 ⇒ x-11/2+y11/4=1 (a=-112, b=114)
(a) (iii) Normal form Divide by sqrt{22+(-4)2}=2sqrt5: -xsqrt5+2ysqrt5=112sqrt5
(a) cosα=-1sqrt5<0, sinα=2sqrt5>0 ⇒ α=116.57^circ
Result
(a) xcos(116.57^circ)+ysin(116.57^circ)=112sqrt5, p=112sqrt5
(b) (i) Slope-intercept 7y=-4x+2 ⇒ y=-tfrac47x+tfrac27 (m=-tfrac47, c=tfrac27)
(b) (ii) Two-intercept 4x+7y=2 ⇒ x1/2+y2/7=1 (a=tfrac12, b=tfrac27)
(b) (iii) Normal form Divide by sqrt{16+49}=sqrt{65}: frac{4x}{sqrt{65}+frac{7y}{sqrt{65}=frac{2}{sqrt{65}
(b) cosα=tfrac{4}{sqrt{65}>0, sinα=tfrac{7}{sqrt{65}>0 ⇒ α=60.26^circ
(b) xcos(60.26^circ)+ysin(60.26^circ)=tfrac{2}{sqrt{65}, p=tfrac{2}{sqrt{65}
(c) (i) Slope-intercept 15y=8x-3 ⇒ y=815x-tfrac15 (m=815, c=-tfrac15)
(c) (ii) Two-intercept 8x-15y=3 ⇒ x3/8+y-1/5=1 (a=tfrac38, b=-tfrac15)
(c) (iii) Normal form Divide by sqrt{64+225}=17: 8x17-15y17=317
(c) cosα=817>0, sinα=-1517<0 ⇒ α=298.07^circ (4th quadrant)
(c) xcos(298.07^circ)+ysin(298.07^circ)=317, p=317
7.2.11.(a) Check whether the two lines 2x + y - 3 = 0 and 4x + 2y + 5 = 0 are parallel, perpendicular, or neither. (b) Check whether the two lines 3y = 2x + 5 and 3x + 2y - 8 = 0 are parallel, perpendicular, or neither. (c) Check whether the two lines 4y + 2x - 1 = 0 and x - 2y - 7 = 0 are parallel, perpendicular, or neither.
(a) Slope of l1 m1=-frac21=-2
(a) Slope of l2 m2=-frac42=-2
Result
(a) m1=m2 ⇒ l1 and l2 are parallel
(b) l1 rearranged 3y=2x+5 ⇒ 2x-3y+5=0
(b) Slope of l1 m1=-2-3=frac23
(b) Slope of l2 m2=-frac32
(b) m1m2=left(frac23right)left(-frac32right)=-1 ⇒ l1,l2 are perpendicular
(c) l1 rearranged 4y+2x-1=0 ⇒ 2x+4y-1=0
(c) Slope of l1 m1=-frac24=-frac12
(c) Slope of l2 m2=-1-2=frac12
(c) m1neq m2, m1m2=-frac14neq-1 ⇒ l1,l2 neither parallel nor perpendicular
7.2.12.Find an equation of the line through (-4, 7) and parallel to the line 2x - 7y + 4 = 0.
Slope of given line m=-2-7=frac27
Formula
y-7=frac27(x-(-4))
7(y-7)=2(x+4) ⇒ 7y-49=2x+8
2x+8-7y+49=0
Result
2x-7y+57=0
7.2.13.Find an equation of the line through (5, -8) and perpendicular to the join of A(-15, -8), B(10, 7).