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Introduction to 3D coordinate geometry · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 18 questions.

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Question 14 Marks
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Answer
Here, A(a, b, c), B(b, c, a), C(c, a, b) $\text{AB}=\sqrt{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2 + (\text{c}-\text{a})^2}$ $=\sqrt{\text{a}^2 +\text{b}^2 -2\text{ab} + \text{b}^2 +\text{c}^2 - 2\text{bc} +\text{c}^2 +\text{a}^2 -2\text{ac}}$ $\text{AB}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab} - 2\text{bc}- 2\text{ac}}$ $\text{BC}=\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2 + (\text{a}-\text{b})^2}$ $=\sqrt{\text{b}^2 +\text{c}^2 - 2\text{bc}+ \text{c}2 + \text{a}^2 -2\text{ca} +\text{a}^2 +\text{b}^2 -2\text{ab}}$ $\text{BC}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab}-2\text{bc} -2\text{ca}}$ $\text{CA}=\sqrt{(\text{a}-\text{c})^2+(\text{b}-\text{a})^2 + (\text{c}-\text{b})^2}$ $=\sqrt{\text{a}^2 +\text{c}^2 - 2\text{ac}+ \text{b}2 + \text{a}^2 -2\text{ab} +\text{b}^2 +\text{c}^2 -2\text{bc}}$ $\text{CA}=\sqrt{2\text{a}^2 + 2\text{b}^2 + 2\text{c}^2 - 2\text{ab}-2\text{bc} -2\text{ca}}$ Since, AB = BC = CA, so $\triangle\text{ABC}$ is an isosceles A.
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Question 24 Marks
Show that the points A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) are the vertices of a rhombus.
Answer
Here, $\text{AB}=\sqrt{(1+1)^2+(3-6)^2+(4-10)^2}$ $=\sqrt{4 +9+ 36}$ $=7\text{ units}$ $\text{BC}=\sqrt{(-1+7)^2+(6-4)^2+(0-7)^2}$ $=\sqrt{36+4+9}$ $=7\text{ units}$ $\text{CD}=\sqrt{(-7+5)^2+(4-1)^2+(7-1)^2}$ $=\sqrt{4 +9+ 36}$ $=7\text{ units}$ $\text{DA}=\sqrt{(-5-1)^2+(1-3)^2+(1-4)^2}$ $=\sqrt{36+4+9}$ $=\sqrt{52}$ $=7\text{ units}$ Since, AB = BC = CD = DA So, ABCD is a rhombus.
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Question 34 Marks
Show that the points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.
Answer
Let A = (0, 7, 10), B = (-1, 6, 6) and C = (-4, 9, 6)
$\text{AB}=\sqrt{(0+1)^2+(7-6)^2+(10-6)^2}$
$=\sqrt{(1)^2+(1)^2+(4)^2}$
$=\sqrt{18}$
$=3\sqrt{2}\text{ units}$
$\text{BC}=\sqrt{(-1+4)^2+(6-9)^2+(6-6)^2}$
$=\sqrt{(3)^2+(3)^2+(0)}$
$=\sqrt{18}$
$=3\sqrt{2}\text{ units}$
$\text{AC}=\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}$
$=\sqrt{(4)^2+(-2)^2+(4)^2}$
$=\sqrt{36}$
$=6\text{ units}$
$(\text{AB})^2+\text{(BC)}^2$
$=(3\sqrt{2})^2+(3\sqrt{2})^2$
$=18+18$
$=36$
$=\text{(AC)}^2$
Also (AB) = (BC)
Hence (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of an isosceles right-angled triangle.
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Question 44 Marks
Using distance formula prove that the following points are collinear: P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)
Answer
P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)
$\text{PQ}=\sqrt{(0-1)^2+(7-4)^2+(-7+5)^2}$
$=\sqrt{(1)^2+(3)^2+(-2)^2}$
$=\sqrt{1+9+4}$
$=\sqrt{14}\text{ units}$
$\text{QR}=\sqrt{(1+1)^2+(4-10)^2+(-5+9)^2}$
$=\sqrt{(2)^2+(-6)^2+(4)^2}$
$=\sqrt{4+36+16}$
$=2\sqrt{14}\text{ units}$
$\text{PR}=\sqrt{(0+1)^2+(7-10)^2+(-7+9)^2}$
$=\sqrt{1^2+(-3)^2+(2)^2}$
$=\sqrt{1+9+4}$
$=\sqrt{14}\text{ units}$
Since, PQ + PR = QR
so, P, Q, R are collinear.
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Question 54 Marks
Find the coordinates of the point which is equidistant from the four points $O(0, 0, 0), A(2, 0, 0), B(0, 3, 0)$ and $C(0, 0, 8)$.
Answer
Let the required point be $P\left(x_1 y_1 z\right)$
Here, $0(0,0,0), A(2,0,0), B(0,3,0), C(0,0,8)$
$\text { Since, }(O P)^2=(P A)^2$
$(x-0)^2+(Y-0)^2+(z-0)^2=(x-2)^2+(y-0)^2+(z-0)^2$
$x^2+y^2+z^2=x^2-4 x+4+y^2+z^2$
$4 x=4$
$X=1$
$(O P)^2=(P B)^2$
$(x-0)^2+(y-0)^2+(z-0)^2=(x-0)^2+(y-3)^2+(z-0)^2$
$x^2+y^2+z^2=x^2+y^2-6 y+9+z^2$
$6 y=9$
$y=\frac{3}{2}$
$(O P)^2=(P C)^2$
$(x-0)^2+(y-0)^2+(z-0)^2=(x-0)^2+(Y-0)^2+(z-8)^2$
$x^2+y^2+z^2=x^2+y^2+z^2-16 z+64$
$16 z=64$
$z=4$
The required point $=\Big(1,\ \frac{3}{2},\ 4\Big)$
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Question 64 Marks
Show that the points (3, 2, 2), (-1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.
Answer
Here, $\text{OA}=\sqrt{(1-3)^2+(3-2)^2+(4-2)^2}$ $=\sqrt{4+1+4}$ $=3\text{ units}$ $\text{OB}=\sqrt{(1+1)^2+(3-1)^2+(4-3)^2}$ $=\sqrt{4+4+1}$ $=3\text{ units}$ $\text{OC}=\sqrt{(1-0)^2+(3-5)^2+(4-6)^2}$ $=\sqrt{1+4+4}$ $=3\text{ units}$ $\text{OD}=\sqrt{(1-2)^2+(3-1)^2+(4-2)^2}$ $=\sqrt{1+4+4}$ $=3\text{ units}$ Since, OA = OC = OD = OB, points A, B, C, O lie on a sphere with centre O. Radius = 3 units.
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Question 74 Marks
Prove that the tetrahedron with vertices at the points O(0, 0, 0), A(0, 1, 1), B(1, 0, 1) and C(1, 1, 0) is a regular one.
Answer
Here, $\text{AB}=\sqrt{(0-1)^2+(1-0)^2+(1-1)^2}$ $=\sqrt{1+1}$ $=\sqrt{2}\text{ units}$ $\text{BC}=\sqrt{(1-1)^2+(0-1)^2+(1-0)^2}$ $=\sqrt{1+1}$ $=\sqrt{2}\text{ units}$ $\text{CA}=\sqrt{(1-0)^2+(1-1)^2+(0-1)^2}$ $=\sqrt{1+0+1}$ $=\sqrt{2}\text{ units}$ $\text{DA}=\sqrt{(0-0)^2+(0-1)^2+(0-1)^2}$ $=\sqrt{1+1}$ $=\sqrt{2}\text{ units}$ $\text{OB}=\sqrt{(0-1)^2+(0-0)^2+(0-1)^2}$ $=\sqrt{1+1}$ $=\sqrt{2}\text{ units}$ $\text{DA}=\sqrt{(0-1)^2+(0-1)^2+(0-0)^2}$ $=\sqrt{1+1}$ $=\sqrt{2}\text{ units}$ Since, OA = OB = OC = AB = BC = CA So, O, A, B, C represent a regular tetrahedron.
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Question 84 Marks
Prove that the point A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.
Answer
Here
$\text{AB}=\sqrt{(1+5)^2+(3-5)^2+(0-2)^2}$
$=\sqrt{36+4+4}$
$=\sqrt{44}$
$=2\sqrt{11}\text{ units}$
$\text{BC}=\sqrt{(-5+9)^2+(5+1)^2+(2-2)^2}$
$=\sqrt{16+36}$
$=\sqrt{52}$
$=2\sqrt{13}\text{ units}$
$\text{CD}=\sqrt{(-9+3)^2+(-1+3)^2+(2-0)^2}$
$=\sqrt{36+4+4}$
$=2\sqrt{11}\text{ units}$
$\text{DA}=\sqrt{(-3-4)^2+(-3-3)^2+0}$
$=\sqrt{16+36}$
$=\sqrt{52}$
$=2\sqrt{13}\text{ units}$
$\text{AC}=\sqrt{(1+9)^2+(3+1)^2+(0-2)^2}$
$=\sqrt{150+16+4}$
$=\sqrt{120}$
$=4\sqrt{5}\text{ units}$
$\text{BD}=\sqrt{(-3+5)^2+(-3-5)^2+(0-2)^2}$
$=\sqrt{4+64+4} $
$=\sqrt{72}$
$=6\sqrt{2}\text{ units}$
Since,
AB = CD and BC = DA
⇒ ABCD is a parallelgram = BD
but, $\text{AC}\neq\text{BD}$
⇒ ABCD is not a rectangles.
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Question 94 Marks
If $A(-2, 2, 3)$ and $B(13, -3, 13)$ are two points. Find the locus of a point $P$ which moves in such a way that $3PA = 2PB$.
Answer
$\text { Let } \mathrm{P} \text { be }\left(\mathrm{x}_1 \mathrm{y} 1 \mathrm{z}\right) \text {, Here, } \mathrm{A}(-2,2,3), \mathrm{B}(13,-3,13) \text { and } 3 P A=2 P B \Rightarrow 3 \sqrt{(x+2)^2(\mathrm{y}-2)^2+(z-3)^2}$
$=2 \sqrt{(\mathrm{x}-13)^2+(\mathrm{y}+3)^2+(z-13)^2} \text { squaring both the sides, }$
$\Rightarrow 9\left[x^2+4 x+4+\mathrm{y}^2+4-4 Y+z^2+9-6 z\right]=$
$4\left[x^2+169-26 x+y^2+9+6 y+z^2+169-26 z\right]$
$\Rightarrow 9 x^2-4 x^2+36 x+104 x+36-676+9 y^2-4 y^2+36-36-36 y-24 y$
$+9 z^2-4 z^2+81-676-54 z+6 y z=0$
$\Rightarrow 5 x^2+5 y^2+5 z 2+140 x-60 y+50 z-1235=0$
$\Rightarrow 5\left(x^2+y^2+z^2\right)+140 x-$
$60 y+50 z-1235=0$
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Question 104 Marks
Using distance formula prove that the following points are collinear: A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
Answer
A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
$\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{(4-5)^2+(-3+7)^2+(-1-6)^2}$
$=\sqrt{(-1)^2+(4)^2+(-7)^2}$
$=\sqrt{1+16+49}$
$=\sqrt{66}\text{ units}$
$\text{BC}=\sqrt{(5-3)^2+(-7-1)^2+(6+8)^2}$
$=\sqrt{(2)^2+(-8)^2+(14)^2}$
$=\sqrt{4+64+196}$
$=\sqrt{264}$
$=2\sqrt{66}\text{ units}$
$\text{AC}=\sqrt{(4-3)^2+(-3-1)^2+(-1+8)^2}$
$=\sqrt{(1)^2+(-4)^2+(7)^2}$
$=\sqrt{1+16+49}$
$=\sqrt{66}\text{ units}$
Since, AC + AB = BC
so, A, B, C are collinear.
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Question 114 Marks
If $A(-2, 2, 3)$ and $B(13, -3, 13)$ are two points. Find the locus of a point P which moves in such a way that $3PA = 2PB$.
Answer
$\text { Let } P \text { be }(\mathrm{x} 1 \mathrm{y} 1 \mathrm{z}) \text {, Here, } A(-2,2,3), B(13,-3,13) \text { and } 3 P A=2 P B \Rightarrow 3 \sqrt{(x+2)^2(y-2)^2+(z-3)^2}$
$=2 \sqrt{(x-13)^2+(y+3)^2+(z-13)^2} \text { squaring both the sides, }$
$\Rightarrow 9\left[x^2+4 x+4+y^2+4-4 Y+z^2+9-6 z\right]=$
$4\left[x^2+169-26 x+y^2+9+6 y+z^2+169-26 z\right]$
$\Rightarrow 9 x^2-4 x^2+36 x+104 x+36-676+9 y^2-4 y^2+36-36-36 y-24 y$
$+9 z^2-4 z^2+81-676-54 z+6 y z=0$
$\Rightarrow 5 x^2+5 y^2+5 z 2+140 x-60 y+50 z-1235=0$
$\Rightarrow 5\left(x^2+y^2+z^2\right)+140 x-$
$60 y+50 z-1235=0$
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Question 124 Marks
Find the locus of $P$ if $P A^2+P B^2=2 k^2$, where $A$ and $B$ are the points $(3,4,5)$ and $(-1,3,-7)$.
Answer
Let $\mathrm{P}(\mathrm{x} 1, \mathrm{y} 1, \mathrm{z})$,
Here, $\mathrm{A}(3,4,5), \mathrm{B}(-1,3,-7) \mathrm{PA}^2+\mathrm{PB}^2=2 \mathrm{k}^2$
$\Rightarrow(\mathrm{x}-3)^2+(\mathrm{y}-4)^2+(\mathrm{z}-5)^2+(\mathrm{x}+1)^2+(\mathrm{y}-3)^2+(2+7)^2$
$=2 k^2$
$\Rightarrow x^2+9-6 x+y^2+16-8 y+z^2+25-10 z+x^2+1+2 x+y^2+9-6 y+z^2+49+14 z=2 k^2$
$\Rightarrow 2 x^2+2 y^2+2 z^2-4 x-14 y+4 z+109=2 k^2$
$\Rightarrow 2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+109-2 k^2=0$
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Question 134 Marks
Show that the plane $a x+b y+c z+d=0$ divides the line joining the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ in the ratio $=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+d}{\text{ax}_2+\text{by}_2+\text{cz}_2+d}.$
Answer
Assume ratio is $\lambda:1$
Plane is ax + by + cz + d = 0
points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$
Assume point of intersection of line and plane is D
$\text{D}=\Big(\frac{\lambda\text{x}_2+\text{x}_1}{\lambda+1},\frac{\lambda\text{y}_2+\text{y}_1}{\lambda+1},\frac{\lambda\text{z}_2+\text{z}_1}{\lambda+1}\Big)$
As D lies on plane, substitute D in plane equation, we get
$\lambda(\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d})+\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}=0$
$\Rightarrow\lambda=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$
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Question 144 Marks
Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelopiped so formed.
Answer
IMAGE
Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.
$\therefore$ PA = Distance between planes PBEC and QDAF
= 5 - 3
= 2
PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and -2, respectively, from the zx-plane.
$\therefore$ PB = 0 - (-2)
= 2
PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.
$\therefore$ PC = 2 - 5
= -3
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Question 154 Marks
Determine the points yz-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1)$.
Answer
Let $Q(0, y, z)$ be the required point.
So
$(A Q)^2=(B Q)^2 \Rightarrow(0-1)^2+(y+1)^2-(z-0)^2=(0-2)+(y+1)^2+(z-2)^2$
$\Rightarrow 1+y^2+1+2 y+z^2=4+y^2+1-2 y+z^2+4-42$
$\Rightarrow 4 y+4 z=7 \ldots(\text { i })$
$(B Q)^2=(C Q)^2 \Rightarrow(0-z)^2+(y-1)^2+(z-2)^2=(0-3)^2+(y-2)^2(2+1)^2$
$\Rightarrow 4+y^2+1-2 y+z^2+4-4 z-9+y^2+4-4 y+z^2+1+2 z$
$\Rightarrow 2 y-6 z=5 \ldots(\text { (ii) }$
$(\mathrm{AQ})^2=(C Q)^2 \Rightarrow(0-1)^2+(y+1)^2+(z-0)^2=(0-3)^2+(y-2)^2(z+1)^2$
$\Rightarrow 1+y^2+2 y+1+z^2=9+y^2-4 y+4+z^2+1+2 z$
$\Rightarrow 6 y-2 z=12 \ldots \text { (iii) }$
Solving equation (i) and (ii) we get
$\text{z}=\frac{-3}{16}$ and $\text{y} = \frac{31}{16}$
Put the value of y and z in equation (iii)
$6y - 2z = 12 = 12$
$6\Big(\frac{31}{16}\Big)-2\Big(\frac{-3}{16}\Big) = 12$
$\frac{186}{16}+\frac{6}{16}=12$
$\frac{192}{16}=12$
$12=12$
LHS = RHS.
so,
$\text{y}=\frac{31}{16},\ \text{z}=\frac{13}{16}$
Required point $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$
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Question 164 Marks
Determine the points xy-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1)$.
Answer
Let the point on $x y$-plane be $P(x, y, 0)$.
Now $P$ is equidistance from $A(1,-1,0), B(2,1,2)$ and $C(3,2,-1)$.
So, AP - BP - CP
Now,
$(A P)^2=(x-1)^2+(y+1)^2+(0-0)^2$
$(B P)^2=(x-2)^2+(y-1)^2+(0-2)^2$
$(C P)^2=(x-3)^2+(y-2)^2+(0+1)^2$
$(A P)^2=(B P)^2 \Rightarrow(x-1)^2+(y+1)^2-(x-2)^2+(y-1)^2+4$
$\Rightarrow x^2+1-2+y^2+1+2 y+z^2-x^2+4-4 x+y^2+1-2 y+4$
$\Rightarrow 2 x+4 y=7 \ldots(i)$
$(B P)^2=(C P)^2 \Rightarrow(x-2)^2+(y-1)^2+4=(x-3)^2+(y-2)^2+1$
$\Rightarrow x^2+4-4 x+y^2+1-2 y+z^2+4=x^2+9-6 x+y^2+4-4 y+1$
$\Rightarrow 2 x+2 y=5 \ldots(\text { ii) }$
$(A P)^2=(C P)^2 \Rightarrow(x-1)^2+(y+1)^2+(x-3)^2+(y-2)^2+1$
$\Rightarrow x^2+1-2 x+y^2+1+2 y=x^2+9-6 x+y^2+4-4 y+1$
$\Rightarrow 4 x+5 y=12 \ldots \text { (iii) }$
Solving equation (i) and (ii) we get
$y=1, x=\frac{3}{2}$
Put x and y in equation (iii)
y = 1, $\text{x} = \frac{3}{2}$
Put x and y in equation (iii)
$4\Big(\frac{3}{2}\Big) + 6(1) = 12$
$12-12$
So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$
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Question 174 Marks
Determine the points zx-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1)$.
Answer
Let $R(x, 0, z)$ be the required point.
So
$(A R)^2=(B R)^2 \Rightarrow(1-x)^2+(-1-0)^2+(0-z)^2=(2-x)+(1-0)^2+(2-z)^2$
$\Rightarrow 1+x^2-2 x+1+z^2=4+x^2-4 x+1+z^2+4 z$
$\Rightarrow 2 x+4 z=7 \ldots(\mathrm{i})$
$(B R)^2=(C R)^2 \Rightarrow(z-z)^2+(1-0)^2+(2-z)^2=(3-x)^2+(2-0)^2(-1-z)^2$
$\Rightarrow 4+x^2-4 x+4+z^2-4 z=9+x^2-6 x+4+1+z^2+2 z$
$\Rightarrow 2 x-6 z=5 \ldots(\text { ii) }$
$(A R)^2=(C R)^2 \Rightarrow(1-x)^2+(1-0)^2+(0-z)^2=(3-x)^2+(2-0)^2+(-1-z)^2$
$\Rightarrow 1+x^2-2 x+1+z^2=9+6 x+4+1+z^2$
$\Rightarrow 4 x-2 z=12 \ldots \text { (iii) }$
Solving equation (i) and (ii) we get
$\text{z}=\frac{1}{5},\ \text{x}=\frac{31}{10}$
Put the value of x and z in equation (iii)
$4x - 2z = 12$
$4\Big(\frac{31}{10}\Big)-2\Big(\frac{1}{5}\Big) = 12$
$\frac{124}{10}+\frac{2}{10}=12$
$\frac{120}{10}=12$
$12=12$
LHS = RHS.
so,
$\text{x}=\frac{31}{10},\ \text{z}=\frac{1}{5}$
Required point $=\Big(\frac{31}{10},\ 0,\ \frac{1}{5}\Big)$
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Question 184 Marks
Using distance formula prove that the following points are collinear: A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
Answer
A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
$\text{AB}=\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}$
$=\sqrt{(4)^2+(-5)^2+(-7)^2}$
$=\sqrt{16+25+49}$
$=\sqrt{90}$
$=3\sqrt{10}\text{ units}$
$\text{BC}=\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}$
$=\sqrt{(-8)^2+(10)^2+(14)^2}$
$=\sqrt{64+100+196}$
$=\sqrt{360}$
$=6\sqrt{10}\text{ units}$
$\text{CA}=\sqrt{(3-7)^2+(-5+10)^2+(1+6)^2}$
$=\sqrt{(-4)^2+(5)^2+(7)^2}$
$=\sqrt{16+25+49}$
$=\sqrt{90}$
$=3\sqrt{10}\text{ units}$
Since, AB + AC = BC
so, A, B and C are collinear.
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