Note: We are going to use the following formula extensively in solving the following problems. The distance between any two points $\displaystyle (x_1, y_1) \text{ and } (x_2, y_2)$ is $\displaystyle = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Notes: If a point is on $\displaystyle x-axis$ , its ordinate is $\displaystyle 0$ , therefore the point on $\displaystyle x-axis$ is taken as $\displaystyle (x,0)$ . Similarly, if the point is on $\displaystyle y-axis$ , its abscissa is $\displaystyle 0$ , therefore the point on $\displaystyle y-axis$ is taken as $\displaystyle (0,y)$ . For details refer to the following lecture notes.

Question 1: Find the distance between the following pairs of points:

$\displaystyle \text{i) } (-3,6) \text{ and } (2, -6)$

$\displaystyle \text{ii) } (-a, -b) \text{ and } (a, b)$

$\displaystyle \text{iii) } (\frac{3}{5}, 2) \text{ and } (-\frac{1}{5}, 1\frac{2}{5})$

$\displaystyle \text{iv) } (\sqrt{3}+1, 1) \text{ and } (0, \sqrt{3})$

$\displaystyle \text{i) } (-3, 6) \text{ and } (2, -6)$

$\displaystyle \text{Distance } = \sqrt{(2+3)^2+(-6-6)^2} = \sqrt{25+144} = \sqrt{169} = 13$

$\displaystyle \text{ii) } (-a, -b) \text{ and } (a, b)$

$\displaystyle \text{Distance } = \sqrt{(a+a)^2+(b+b)^2} = \sqrt{4a^2+4b^2} = 2\sqrt{a^2+b^2}$

$\displaystyle \text{iii) } (\frac{3}{5}, 2)$ and ( $\displaystyle -\frac{1}{5}, 1\frac{2}{5})$

$\displaystyle \text{Distance } = \sqrt{(-\frac{1}{5}-\frac{3}{5})^2+(1\frac{2}{5}-2)^2} = \sqrt{\frac{16}{25}+\frac{9}{25}} = 1$

$\displaystyle \text{iv) } (\sqrt{3}+1, 1) \text{ and } (0, \sqrt{3})$

$\displaystyle \text{Distance } = \sqrt{(0-\sqrt{3}-1)^2+(\sqrt{3}-1)^2}$

$\displaystyle = \sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$

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Question 2: Find the distance between the origin and the points:

$\displaystyle \text{i) } (-8, 6)$

$\displaystyle \text{ii) } (-5, -12)$

$\displaystyle \text{iii) } (8, -15)$

$\displaystyle \text{i) } (0,0) \text{ and } (-8, 6)$

$\displaystyle \text{Distance } = \sqrt{(-8-0)^2+(6-0)^2} = \sqrt{64+36} = \sqrt{100} = 10$

$\displaystyle \text{ii) } (0,0) \text{ and } (-5, -12)$

$\displaystyle \text{Distance } = \sqrt{(-5-0)^2+(-12-0)^2} = \sqrt{25+144} = \sqrt{169} = 13$

$\displaystyle \text{iii) } (0,0) \text{ and } (8, -15)$

$\displaystyle \text{Distance } = \sqrt{(8-0)^2+(-15-0)^2} = \sqrt{64+225} = \sqrt{289} = 17$

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Question 3: The distance between point $\displaystyle (3, 1) \text{ and } (0, x)$ is $\displaystyle 5$ . Find $\displaystyle x$

$\displaystyle (3, 1) \text{ and } (0, x)$

$\displaystyle \text{Distance } \sqrt{(0-3)^2+(x-1)^2} = 5$

$\displaystyle \sqrt{9+(x-1)^2} = 5$

$\displaystyle (x-1)^2=25-9=16$

$\displaystyle x^2+1-2x=16$

$\displaystyle x^2-2x-15=0$

$\displaystyle (x-5)(x+3)=0 \Rightarrow x = 5 or -3$

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Question 4: Find the coordinate of the point on $\displaystyle x-axis$ which are at a distance of $\displaystyle 17$ units from the point $\displaystyle (11, 8)$

$\displaystyle (x,0) \text{ and } (11, -8)$

$\displaystyle \text{Distance } \sqrt{(11-x)^2+(-8-0)^2} =17$

$\displaystyle (11-x)^2+64=289$

$\displaystyle (11-x)^2=225$

$\displaystyle x^2-22x-104=0$

$\displaystyle (x-26)(x+4)=0 \Rightarrow x = 26 \ or \ -4$

Therefore the points are $\displaystyle (26,0) \text{ and } (-4,0)$

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Question 5: Find the coordinate of the point on $\displaystyle y-axis$ which are at a distance of $\displaystyle 10$ units from the point $\displaystyle (11, -8)$

$\displaystyle (0,y) \text{ and } (-8, 4)$

$\displaystyle \text{Distance } \sqrt{(-8-0)^2+(4-y)^2} =10$

$\displaystyle 64+(4-y)^2=100$

$\displaystyle (4-y)^2=36$

$\displaystyle y^2-8y-20=0$

$\displaystyle (y-10)(y+2)=0 \Rightarrow y = 10 \ or \ -2$

Hence the points could be $\displaystyle (0, 10) \text{ and } (0, -2)$

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Question 6: A point $\displaystyle A$ is at a distance of $\displaystyle \sqrt{10}$ units from the point $\displaystyle (4, 3)$ . Find the coordinates of the point $\displaystyle A$ if its ordinate is twice its abscissa.

$\displaystyle (2a,a) \text{ and } (4,3)$

$\displaystyle \text{Distance } \sqrt{(4-2a)^2+(3-a)^2} =\sqrt{10}$

$\displaystyle (4-2a)^2+(3-a)^2=10$

$\displaystyle 5a^2-22a-15=0$

$\displaystyle (a-5)(5a+3)=0 \Rightarrow a =5 \ or \ -\frac{3}{5}$

$\displaystyle \text{Hence the points could be } (10,5) \text{ and } (-\frac{6}{5}, -\frac{3}{5})$

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Question 7: A point $\displaystyle P (2, 1)$ is equidistant from the point $\displaystyle (a, 7) \text{ and } (3, a)$ . Find $\displaystyle a$

$\displaystyle P(2, -1)$ is equidistant from $\displaystyle (a, 7) \text{ and } (-3, 9)$

Therefore $\displaystyle \sqrt{(a-2)^2+(7+1)^2} = \sqrt{(-3-2)^2+(a+1)^2}$

$\displaystyle (a-2)^2+64=25+(a+1)^2$

$\displaystyle a^2+4-4a+64=25+a^2+1+2a$

$\displaystyle 6a=42 \Rightarrow a =7$

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Question 8: What point on $\displaystyle x-axis$ is equidistant from the point $\displaystyle (7, 6) \text{ and } (-3, 4)$

Let the point be $\displaystyle P(x,0)$ . Therefore

$\displaystyle \sqrt{(7-x)^2+(6-0)^2} = \sqrt{(-3-x)^2+(4-0)^2}$

$\displaystyle (7-x)^2+36=(-3-x)^2+16$

$\displaystyle 49+x^2-14x+36=9+x^2+6x+16$

$\displaystyle 20x=60 \Rightarrow x = 3$

Therefore the point is $\displaystyle (3,0)$

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Question 9: What point on $\displaystyle y-axis$ is equidistant from the point $\displaystyle (5, 2) \text{ and } (-4, 3)$

Let the point be $\displaystyle P (0,y)$ . Therefore

$\displaystyle \sqrt{(5-0)^2+(2-y)^2} = \sqrt{(-4-0)^2+(3-y)^2}$

$\displaystyle 25+(2-y)^2=16+(3-y)^2$

$\displaystyle 9+4+y^2-4y=9+y^2-6y$

$\displaystyle 2y=-4 \Rightarrow y = -2$

Hence the point is $\displaystyle (0, -2)$

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Question 10: A point $\displaystyle P$ lies on $\displaystyle x-axis$ and another point $\displaystyle Q$ lies on $\displaystyle y-axis$ . Write the ordinate of point $\displaystyle P$ , abscissa of point $\displaystyle Q$ . If the abscissa of point $\displaystyle P$ is $\displaystyle -12$ and ordinate of point $\displaystyle Q$ is $\displaystyle -16$ . Calculate the length of the line segment $\displaystyle PQ$

Let $\displaystyle P(x, 0) \text{ and } Q (0, y)$ . Given $\displaystyle P(-12, 0) \text{ and } Q (0, -16)$

$\displaystyle \text{Distance } \sqrt{(0+12)^2+(-16-0)^2} = \sqrt{144+256} = \sqrt{400} = 20$

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Question 11: Show that the points $\displaystyle P(0, 5), Q(5, 10) \text{ and } R(6, 3)$ are the vertices of an isosceles triangle.

$\displaystyle P(0, 5), Q(5, 10) \text{ and } R(6, 3)$

$\displaystyle PQ = \sqrt{(5-0)^2+(10-5)^2} = \sqrt{25+25} = \sqrt{50}$

$\displaystyle PR = \sqrt{(6-0)^2+(3-5)^2} = \sqrt{36+4} = \sqrt{40}$

$\displaystyle QR = \sqrt{(6-5)^2+(3-10)^2} = \sqrt{1+49} = \sqrt{50}$

Therefore two sides $\displaystyle PQ \text{ and } QR$ are equal which makes it an isosceles triangle.

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Question 12: Prove that the points $\displaystyle P (0, -4), Q (6, 2), R (3, 5) \text{ and } S (-3, -1)$ are the vertices of the rectangle $\displaystyle PQRS$

$\displaystyle P (0, -4), Q (6, 2), R (3, 5) \text{ and } S (-3, -1)$

$\displaystyle PQ = \sqrt{(6-0)^2+(2+4)^2} = \sqrt{36+36} = \sqrt{72}$

$\displaystyle QR = \sqrt{(3-6)^2+(5-2)^2} = \sqrt{9+9} = \sqrt{18}$

$\displaystyle RS = \sqrt{(-3-3)^2+(-1-5)^2} = \sqrt{36+36} = \sqrt{72}$

$\displaystyle PS = \sqrt{(-3-0)^2+(-1+4)^2} = \sqrt{9+9} = \sqrt{18}$

Therefore $\displaystyle PQ=RS \text{ and } QR=PS$ .

Hence it is a rectangle.

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Question 13: Prove that the points $\displaystyle A (1, -3), B (-3, 0) \text{ and } C (4, 1)$ are the vertices of an isosceles triangle. Find the area of the triangle.

$\displaystyle A (1, -3), B (-3, 0) \text{ and } C (4, 1)$

$\displaystyle AB = \sqrt{(-3-1)^2+(0+3)^2} = \sqrt{16+9} = \sqrt{25} = 5$

$\displaystyle AC = \sqrt{(4-1)^2+(1+3)^2} = \sqrt{16+9} = \sqrt{25} = 5$

$\displaystyle BC = \sqrt{(4+3)^2+(1-0)^2} = \sqrt{49+1} = \sqrt{50}$

For this to be a right angled triangle we should have $\displaystyle AB^2+AC^2=BC^2$

$\displaystyle AB^2+AC^2 = (5)^2+(5)^2= (50)^2 = BC^2$ . Hence proved that it is a right angled triangle.

$\displaystyle \text{Area } = \frac{1}{2} \times 5 \times 5 = 12.5 \text{ sq. units. }$

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Question 14: Show that the points $\displaystyle A (5, 6), B (1, 5), C (2, 1) \text{ and } D (6, 2)$ are the vertices of the square $\displaystyle ABCD$

$\displaystyle A (5, 6), B (1, 5), C (2, 1) \text{ and } D (6, 2)$

$\displaystyle AB = \sqrt{(1-5)^2+(5-6)^2} = \sqrt{16+1} = \sqrt{17}$

$\displaystyle BC = \sqrt{(2-1)^2+(1-5)^2} = \sqrt{1+16} = \sqrt{17}$

$\displaystyle CD = \sqrt{(6-2)^2+(2-1)^2} = \sqrt{16+1} = \sqrt{17}$

$\displaystyle DA = \sqrt{(6-5)^2+(2-6)^2} = \sqrt{1+16} = \sqrt{17}$

Therefore $\displaystyle AB=BC=CD=DA$ .

Hence it is a square.

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Question 15: Show that $\displaystyle (-3, -2), (-5, -5), (2, -3) \text{ and } (4, 4)$ are the vertices of a rhombus.

$\displaystyle (-3, -2), (-5, -5), (2, -3) \text{ and } (4, 4)$

$\displaystyle AB = \sqrt{(-5+3)^2+(-5+2)^2} = \sqrt{4+9} = \sqrt{13}$

$\displaystyle BC = \sqrt{(2+5)^2+(-3+5)^2} = \sqrt{49+4} = \sqrt{53}$

$\displaystyle CD = \sqrt{(4-2)^2+(4+3)^2} = \sqrt{4+49} = \sqrt{53}$

$\displaystyle DA = \sqrt{(4+3)^2+(4+2)^2} = \sqrt{49+36} = \sqrt{83}$

Therefore $\displaystyle BC=CD$ .

Two sides are equal and the other two sides are of different length. Hence it is a rhombus.

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Question 16: Points $\displaystyle A (-3, -2), B (-6, a), C (-3, -4) \text{ and } D (0, -1)$ are the vertices of a quadrilateral $\displaystyle ABCD$ . Find a if a is negative and $\displaystyle AB=CD$

$\displaystyle A (-3, -2), B (-6, a), C (-3, -4) \text{ and } D (0, -1)$

$\displaystyle AB = \sqrt{(-6+3)^2+(a+2)^2} = \sqrt{9+(a+2)^2}$

$\displaystyle CD = \sqrt{(0+3)^2+(-1+4)^2} = \sqrt{9+9} = \sqrt{18}$

Therefore $\displaystyle \sqrt{18}= \sqrt{9+(a+2)^2}$

$\displaystyle a^2+4+4a+9=18$

$\displaystyle a^2+4a-5=0$

$\displaystyle (a+5)(a-1)=0 \Rightarrow a = -5 \ or \ 1$ . Hence $\displaystyle a = -5$ as it is negative.

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Question 17: The vertices of a triangle are $\displaystyle (5, 1), (11, 1) \text{ and } (11, 9)$ . Find the coordinates of the circumcenter of the triangle.

$\displaystyle (5, 1), (11, 1) \text{ and } (11, 9)$ are the points

Let the coordinates of the circumcenter $\displaystyle = (x, y)$

Therefore $\displaystyle \sqrt{(x-5)^2+(y-1)^2} = \sqrt{(x-11)^2+(y-1)^2} = \sqrt{(x-11)^2+(y-9)^2}$

Hence $\displaystyle {(x-5)^2+(y-1)^2} = {(x-11)^2+(y-1)^2} = {(x-11)^2+(y-9)^2}$

Therefore equation 1:

$\displaystyle {(x-5)^2+(y-1)^2} = {(x-11)^2+(y-1)^2}$

$\displaystyle x^2+25-10x+y^2+1-2y=x^2+121-22x+y^2+1-2y$

$\displaystyle 26-10x-2y=122-22x-2y$

$\displaystyle 12x=96 \Rightarrow x = 8$

Also equation 2:

$\displaystyle {(x-11)^2+(y-1)^2} = {(x-11)^2+(y-9)^2}$

$\displaystyle y^2+1-2y=y^2+81-18y$

$\displaystyle 16y=80 \Rightarrow y = 5$

Hence the coordinates of the circumcenter is $\displaystyle (8,5)$

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Question 18: Given $\displaystyle A (3, 1) \text{ and } B (0, y-1)$ . Find $\displaystyle y$ if $\displaystyle AB=5$

$\displaystyle A (3, 1) \text{ and } B (0, y-1)$

$\displaystyle AB = \sqrt{(0-3)^2+(y-1-1)^2} = \sqrt{9+y^2}$

Therefore $\displaystyle 9+y^2=25 \Rightarrow y^2=16 \Rightarrow y = 4, -4$

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Question 19: Given $\displaystyle A = (x+2, -2) \text{ and } B=(11, 6)$ . Find $\displaystyle x$ if $\displaystyle AB=17$

$\displaystyle A = (x+2, -2) \text{ and } B=(11, 6)$

$\displaystyle AB = \sqrt{(11-x-2)^2+(6+2)^2} = \sqrt{289} = 17$

$\displaystyle (9-x)^2+64=289$

$\displaystyle (9-x)^2=225$

$\displaystyle x^2+81-18x-225=0$

$\displaystyle x^2-18x-144=0$

$\displaystyle (x+6)(x-24)=0 \Rightarrow x = -6 \ or \ 24$

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Question 20: The center of the circle is $\displaystyle (2x-1, 3x+1)$ . Find $\displaystyle x$ if the circle passes through $\displaystyle (-3, -1)$ and the length of the diameter is $\displaystyle 20$ units.

Diameter $\displaystyle = 20$ units i.e.Radius $\displaystyle = 10$ units

$\displaystyle \sqrt{(-3-2x+1)^2+(-1-3x-1)^2} = 10$

$\displaystyle (-2x-2)^2+(-3x-2)^2=100$

$\displaystyle 4x^2+4+8x+9x^2+4+12x=100$

$\displaystyle 13x^2+20x-92=0 \Rightarrow x = 2, -3.539$

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Question 21: The length of the line $\displaystyle PQ$ is $\displaystyle 10$ units and the coordinates of $\displaystyle P$ are $\displaystyle (2, -3)$ , calculate the coordinates of point $\displaystyle Q$ , if its abscissca is $\displaystyle 10$

$\displaystyle PQ=10, P(2, -3)$ and Let $\displaystyle Q(10, y)$

$\displaystyle \sqrt{(10-2)^2+(y+3)^2} = 10$

$\displaystyle 8^2+(y+3)^2=100$

$\displaystyle 64+y^2+9+6y=100$

$\displaystyle y^2+6y-27=0$

$\displaystyle (y+9)(y-3)=0 \Rightarrow y = -9, 3$

Hence the points $\displaystyle Q$ could be $\displaystyle (10, -9) \ or \ (10,3)$

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Question 22: Point $\displaystyle P (2, -7)$ is the center of the circle with radius $\displaystyle 13$ units, $\displaystyle PT$ is perpendicular to chord $\displaystyle AB \text{ and } T=(-2, -4)$ . Calculate the length of $\displaystyle \text{i) } AT$ $\displaystyle \text{ii) } AB$

$\displaystyle P(2, -7)$ ; Radius $\displaystyle = 13$ units; $\displaystyle T(-2,-4)$

$\displaystyle PT = \sqrt{(-2-2)^2+(-4+7)^2} = \sqrt{25} = 5$

Therefore $\displaystyle AT=\sqrt{13^2-5^2}=\sqrt{169-25} = 12$

$\displaystyle AB = 2 \times AT = 2 \times 12 = 24 units.$

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Question 23: Calculate the distance between the two points $\displaystyle P(2, 2) \text{ and } Q (5, 4)$ , correct to three significant figures. [1990]

$\displaystyle P(2, 2) \text{ and } Q (5, 4)$

$\displaystyle \text{Distance } = \sqrt{(5-2)^2+(4-2)^2} = \sqrt{9+4} = \sqrt{13} = 3.61$

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Question 24: Calculate the distance between $\displaystyle A(7, 3) \text{ and } B$ on the $\displaystyle x-axis$ whose abscissa is $\displaystyle 11$ [1997]

$\displaystyle A(7, 3) \text{ and } B(11,0)$

$\displaystyle \text{Distance } = \sqrt{(11-7)^2+(0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$

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Question 25: Calculate the distance between $\displaystyle A(5, -3) \text{ and } B$ on the $\displaystyle y-axis$ whose ordinate is $\displaystyle 9$

$\displaystyle A(5, -3) \text{ and } B (0,9)$

$\displaystyle \text{Distance } = \sqrt{(0-5)^2+(9+3)^2} = \sqrt{25+144} = \sqrt{169} = 13$

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Question 26: Find the point on $\displaystyle y-axis$ whose distance from the point $\displaystyle A(6, 7) \text{ and } B(4, -3)$ are in the ratio of $\displaystyle 1:2$

Let point be $\displaystyle P(0,y)$

$\displaystyle A(6, 7) \text{ and } B(4, -3)$

$\displaystyle \frac{AP}{PB} = \frac{1}{2} = \frac{\sqrt{(0-6)^2+(y-7)^2}}{\sqrt{(0-4)^2+(y+3)^2}}$

$\displaystyle \sqrt{16+(y+3)^2}=2\sqrt{36+(y-7)^2}$

$\displaystyle 16+(y+3)^2=4[36+(y-7)^2]$

$\displaystyle 16+y^2+9+6y=4(36+y^2+49-14y)$

$\displaystyle y^2+25+6y=340+4y^2-56y$

$\displaystyle 3y^2-62y+315=0 \Rightarrow y = 9 or 11.67$

Hence points could be $\displaystyle (0,9) \ or \ (0, 11.67)$

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Question 27: The distance of point $\displaystyle P(x, y)$ from the points $\displaystyle A(1, -3) \text{ and } B(-2, 2)$ are in the ratio $\displaystyle 2:3$ . Show that: $\displaystyle 5x^2+5y^2-34x+70y+58=0$

$\displaystyle P(x, y) A(1, -3) \text{ and } B(-2, 2)$

$\displaystyle \frac{PA}{PB} = \frac{2}{3}$

$\displaystyle \frac{\sqrt{(1-x)^2+(-3-y)^2}}{\sqrt{(-2-x)^2+(2-y)^2}} = \frac{2}{3}$

$\displaystyle 9[(1-x)^2+(-3-y)^2] = 4[(-2-x)^2+(2-y)^2]$

$\displaystyle 9[1+x^2-2x+9+y^2+6y]=4[4+x^2+4x+4+y^2-4y]$

$\displaystyle 9[x^2-2x+10+y^2+6y]=4[x^2+4x+8+y^2-4y]$

$\displaystyle 5x^2+5y^2-34x+70y+58=0$ . Hence proved.

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Question 28: The point $\displaystyle A(3, 0), B(a, -2) \text{ and } C(4, -1)$ are vertices of triangle $\displaystyle ABC$ right angles at vertex $\displaystyle A$ . Find the value of $\displaystyle a$

$\displaystyle A(3, 0), B(a, -2) \text{ and } C(4, -1)$

$\displaystyle AB = \sqrt{(a-3)^2+(-2-0)^2} = \sqrt{4+(a-3)^2}$

$\displaystyle AC = \sqrt{(4-3)^2+(-1-0)^2} = \sqrt{1+1} = \sqrt{2}$

$\displaystyle BC = \sqrt{(4-a)^2+(-1+2)^2} = \sqrt{1+(4-a)^2}$

$\displaystyle AB^2+AC^2=BC^2$

$\displaystyle 4+(a-3)^2+2=1+(4-a)^2$

$\displaystyle 4+a^2+9-6a+2=1+16+a^2-8a$

$\displaystyle 2a=2 \Rightarrow a = 1$