Equation of a Line — ICSE Class 10 Maths Important Questions

Answer: (a) $2$2.

$m=\dfrac{7-3}{4-2}=\dfrac{4}{2}=2.$m=7-3/4-2=4/2=2.

Answer: (a) $-\tfrac23$-23.

$3y=-2x+6\Rightarrow y=-\dfrac23x+2$3y=-2x+6 y=-23x+2, so the slope is $-\dfrac23.$-23.

Answer: (c) $60^\circ$60^.

$y=\sqrt3\,x+2$y=3\,x+2, so the slope is $m=\tan\theta=\sqrt3$m==3, which gives $\theta=60^\circ.$=60^.

Answer: (a) $-\tfrac13$-13.

Slope of the join $=\dfrac{8-2}{3-1}=3$=8-2/3-1=3. Perpendicular slope $=-\dfrac1{3}.$=-13.

Answer: (a) The first line has slope $2$2; the second is $y=2x+7$y=2x+7, slope $2$2. Equal slopes mean the lines are parallel, so R correctly explains A.

$2y=4x+6\Rightarrow y=2x+3.$2y=4x+6 y=2x+3.

Slope $=2$=2 and $y$y-intercept $=3.$=3.

$y-y_1=m(x-x_1)$y-y_1=m(x-x_1): $y-(-3)=-4(x-2)\Rightarrow y+3=-4x+8\Rightarrow 4x+y-5=0.$y-(-3)=-4(x-2) y+3=-4x+8 4x+y-5=0.

Slope $m=\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac12.$m=2-4/-1-3=-2/-4=12.

$y-4=\dfrac12(x-3)\Rightarrow 2y-8=x-3\Rightarrow x-2y+5=0.$y-4=12(x-3) 2y-8=x-3 x-2y+5=0.

$2x-3y=6\Rightarrow y=\dfrac23x-2$2x-3y=6 y=23x-2, slope $\dfrac23$23.

Perpendicular slope $=-\dfrac32.$=-32.

$y-3=-\dfrac32(x+2)\Rightarrow 2y-6=-3x-6\Rightarrow 3x+2y=0.$y-3=-32(x+2) 2y-6=-3x-6 3x+2y=0.

Let the equal intercepts be $a$a $(a\ne0)$(a≠0). Then the line passes through $(a,0)$(a,0) and $(0,a)$(0,a), so its slope is
$m=\dfrac{a-0}{0-a}=-1.$m=a-0/0-a=-1.

It passes through $(4,3)$(4,3): $y-3=-1(x-4)\Rightarrow y-3=-x+4.$y-3=-1(x-4) y-3=-x+4.

The equation is $x+y=7.$x+y=7. (Check: putting $y=0$y=0 gives $x=7$x=7, and putting $x=0$x=0 gives $y=7$y=7, so both intercepts are $7$7.)

(a) Median through $A$A. It joins $A$A to the mid-point $D$D of $BC$BC:
$D=\left(\dfrac{1+5}{2},\dfrac{-1+1}{2}\right)=(3,0).$D=(1+5/2,-1+1/2)=(3,0).

Slope of $AD=\dfrac{0-3}{3-(-1)}=-\dfrac34.$AD=0-3/3-(-1)=-34.

$y-3=-\dfrac34(x+1)\Rightarrow 4y-12=-3x-3\Rightarrow 3x+4y-9=0.$y-3=-34(x+1) 4y-12=-3x-3 3x+4y-9=0.

(b) Altitude through $A$A. It is perpendicular to $BC$BC.

Slope of $BC=\dfrac{1-(-1)}{5-1}=\dfrac12$BC=1-(-1)/5-1=12, so the slope of the altitude $=-2.$=-2.

$y-3=-2(x+1)\Rightarrow y-3=-2x-2\Rightarrow 2x+y-1=0.$y-3=-2(x+1) y-3=-2x-2 2x+y-1=0.

(i) Slope of $AB=\dfrac{0-4}{3-1}=-2.$AB=0-4/3-1=-2. Using $A(1,4)$A(1,4): $y-4=-2(x-1)\Rightarrow 2x+y=6.$y-4=-2(x-1) 2x+y=6.

(ii) Mid-point of $AB=\left(\dfrac{1+3}{2},\dfrac{4+0}{2}\right)=(2,2).$AB=(1+3/2,4+0/2)=(2,2). Perpendicular slope $=\dfrac12.$=12.

$y-2=\dfrac12(x-2)\Rightarrow 2y-4=x-2\Rightarrow x-2y+2=0.$y-2=12(x-2) 2y-4=x-2 x-2y+2=0.

(iii) Put $(0,1)$(0,1): $0-2(1)+2=0$0-2(1)+2=0, which holds, so $(0,1)$(0,1) lies on the perpendicular bisector.

(i) Slope of $AB=\dfrac{3-(-1)}{6-2}=\dfrac{4}{4}=1.$AB=3-(-1)/6-2=4/4=1. Line $L$L has slope $1$1 through $P(-2,3)$P(-2,3): $y-3=1(x+2)\Rightarrow y=x+5.$y-3=1(x+2) y=x+5.

(ii) $x$x-axis ($y=0$y=0): $0=x+5\Rightarrow x=-5$0=x+5 x=-5, point $(-5,0).$(-5,0). $y$y-axis ($x=0$x=0): $y=5$y=5, point $(0,5).$(0,5).

(iii) Solve $y=x+5$y=x+5 and $2x+y=4$2x+y=4: $2x+(x+5)=4\Rightarrow 3x=-1\Rightarrow x=-\dfrac13,\ y=x+5=\dfrac{14}{3}.$2x+(x+5)=4 3x=-1 x=-13, y=x+5=14/3. The point is $\left(-\dfrac13,\dfrac{14}{3}\right).$(-13,14/3).

(i) Slope $=\dfrac{10-2}{4-0}=\dfrac{8}{4}=2.$=10-2/4-0=8/4=2.

(ii) The $y$y-intercept is $2$2, so $y=2x+2.$y=2x+2.

(iii) At $x=3$x=3: $y=2(3)+2=8$y=2(3)+2=8, so $(3,8)$(3,8) does lie on the road.

(iv) A parallel road has slope $2$2; through $(0,-1)$(0,-1) its equation is $y=2x-1.$y=2x-1.